Mechanics of Engineering. 



comprising 

Statics and Dynamics of Solids ; the Mechanics of the 

Materials of Construction, or Strength and Elasticity 

of Beams, Columns, Shafts, Arches, etc. ; and the 

Principles of Hydraulics and Pneumatics, 

with Applications. 



FOR USE IN TECHNICAL SCHOOLS. 



IRVING P. CHURCH, C.E., 

Professor of Applied Mechanics, College of Civil Engineering, 
Cornell University. 



EIGHTH EDITION OF " STATICS AND DYNAMICS." 
SEVENTH EDITION OF " MECHANICS AND MATERIALS.** 
FIFTH EDITION OF " MECHANICS OF FLUIDS." 



THIRD THOUSAND. 

NEW YORK: 

JOHN WILEY & SONS, 

53 East Tenth Street. 

1895. 



<K^ fc 






Copyright, 1890, 

BY 

Seving P. Church. 



jSlectrotyped bg 

jDrtjmjiond & NWJ, 

New York 



Printed by 

Ferris Bros., 

New York. 



/^-^5£ 



PREFACE. 



For the engineering student, pursuing the study of Applied Mechan< 
ics as part; of his professional training, and not as additional mathe- 
mathical culture, not only is a thoroughly systematic, clear, and 
consistent treatment of the subject quite essential, but one which pre- 
sents the quantities and conceptions involved in as practical and con- 
crete a form as possible, with all the aids of the printer's and engraver's 
arts ; and especially one which, besides showing the derivation of 
formulae from principles, illustrates, inculcates, and lays stress on 
correct numerical substitution and the consistent and proper use of 
units of measurement; for without this no reliable results can be 
reached, and the principal object of these formulas is frustrated. 

With these requirements in view, and aided by the experience of ten 
years in teaching the Mechanics of Engineering at this institution, the 
writer has been led to prepare the present work, in which attention is 
called to the following features : 

The diagrams are very numerous (about one to every page ; an appeal 
to the eye is often worth a page of verbal description). 

The symbols for distances, angles, forces, etc., used in the algebraic 
work are, as far as possible, inserted directly in the diagrams, render- 
ing the latter full and explicit, and thus saving time and mental effort 
to the student. In problems in Dynamics three kinds of arrows are 
used to distinguish forces, velocities, and accelerations, respectively, 
and thus to prevent confusion of ideas. 

Illustrations and examples of a practical nature, both algebraic and 
numerical, are of frequent occurrence. 

Formulae are divided into two classes ; those (homogeneous) admit- 
ting of the use of any system of units whatever for measurements of 
force, space, mass, and time, in numerical substitution ; and those 
which are true for specified units only. Attention is repeatedly di- 
rected to the matter of correct numerical substitution, especially in 
Dynamics, where time and mass, as well as force and space, are among 
the quantities considered. The importance, in this connection, of 
frequent mention of the quality of the various kinds of quantity em- 
ployed, is also recognized, and a corresponding phraseology adopted. 

The definition of force (§ 3) is made to include and illustrate Newton's 
law of action and reaction, the misconception of which leads to such 
lengthy discussions in technical journals every few years. 

In the matter of " Centrifugal force," the artificial method, so com- 
monly adopted, of regarding a particle moving uniformly in a circle 
as in equilibrium, i. e., acted on by a balanced system of forces, one of 
which is the "Centrifugal force," has been avoided, as being at vari- 
ance with a system of Mechanics founded on Newton's laws, according 
to the first of which a particle moving in any other than a straight line 
cannot be in equilibrium. In such a system of Mechanics nothing can 



iv PREFACE. 

be recognized as a force which is not a definite pull, pnsh, pressure, 
rub, attraction or repulsion, of one body upon, or against, another. 

It is true that the artificial nature of the method referred to is in 
some text-books fully explained in the context, (in Goodeve's Steam 
Engine, for instance, in treating the governor ball,) but is too often not 
mentioned at all, so that the student risks being led into error in 
attempting kindred problems by what would then seem to him correct 
methods. 

The general theorem of Work and Energy in machines is developed 
gradually by definite and limited steps, in preference to giving a single 
demonstration which, from its generality, might be too vague and ab- 
struse to be readily grasped by the student. 

In the use of the Calculus, (in the elements of which the student is 
supposed to have had the training usually given in technical schools by 
the end of the second year) the integral sign is always used to indicate 
summation (except on p. 357) while the name of anti-derivative of a 
given function (of one variable) has been adopted for that function whose 
derivative, or differential co-efficient, is the given function (see §253. ) 

The signs -\ and || are used for perpendicular and parallel, respect- 
ively. 

In Torsion and Flexure of Beams, the well worn and simple theories 
of Navier have been thought sufficient for establishing practical for- 
mulae for safe loads and deflections of beams and shafts ; and promi. 
nence has been given to the methods of designing the cross-sections 
and riveting of built-beams and plate -girders, forming the basis of the 
tables and rules usually given in the pocket-books of our iron and steel 
manufacturers. 

The analytical treatment of the continuous girder is not presented in 
the general case, preference being given to the graphic method by 
Mohr, as greatly superior in simplicity, directness, and interest. For 
similar reasons the graphics of the arch of masonry is to be preferred to 
the analytical chapter on Linear Arches, whose insertion is chiefly a 
concession to the mathematical student, as are also §§ 119, 198, 234, 235, 
284, 265, 266, 284, 287, 291, and 297. 

The graphics of curved beams or arch ribs is made to precede that of 
the straight girder, since the treatment of the latter as a particular case 
of the former is then a comparatively simple matter. Hence Prof. 
Eddy's methods* (inserted by his kind permission) for the arch rib of 
hinged ends, and also that of fixed ends, are presented as special geo- 
metrical devices, instead of being based on Prof. Eddy's general theorem 
(involving a straight girder of the same section and mode of support). 

Acknowledgment is also due Prof. Burr and Prof. Robinson, for 
their cordial consent to the use of certain items and passages from 
their works ; (see §§206, 212, 220, and 297.) 

* See pp. 14 and 25 of " Researches in Graphical Statics." by Prof. H. T. Eddy, 
C.E., Ph. D.; published by D. Van Nostrand, New York, 1878 : reprinted from Van Nos- 
trand's Magazine for 1877. 



PREFACE. t 

Advantage has been taken of the results of the most recent experimental 
investigations in Hydraulics in assigning values of the numerous coeffi- 
cients necessary in this science. The researches of Messrs. Fteley and 
Stearns in 1880 and of M. Bazin in 1887 on the flow of water over weirs, 
and of Mr. Clemens Herschel in testing his invention the " Venturi Water- 
meter," are instances in point; as also some late experiments on the 
transmission of compressed air and of natural gas, and Mr. Freeman's 
extensive investigations in the Hydraulics of Fire-streams and resistance 
of Fire-hose, p. 832. (See Transac. Am. Soc. Civ. Eng. for Nov. 1889.) 

In dealing with fluid tension care has been taken to use the absolute 
pressure and not simpty the excess over atmospheric, thus avoiding the 
occurrence of the term " negative pressure;" this precaution being specially 
necessary in the treatment of gaseous fluids. 

Though space has forbidden dealing at any great length with the action of 
fluid motors, sufficient matter is given in treating of the mode of working 
of steam, gas, and hot-air engines, air- compressors, and pumping-engines, 
together with numerical examples, to be of considerable advantage, it is 
thought, to students not making a specialty of mechanical engineering. 

Special acknowledgment is due to Col. J. T. Fanning, the well-known 
author of " Hydraulic and Water-supply Engineering," for his consent to 
the use of an abridgment of the table of coefficients for friction of water 
in pipes, given in that work; and to Prof. C. L. Crandall, of this univer- 
sity, for permission to incorporate the chapter on Retaining- Walls. 

References to original research in the Hydraulic Laboratory of the 
Civil Engineering Department at this institution will be found on pp. 694 
and 729. 

Cornell University, Ithaca, N. Y., 
January, 1890. 



TABLE OF CONTENTS. 

{MECHANICS OF SOLIDS.] 



PRELIMINARY CHAPTER. 

PAGE 

§§ l-15a. Definitions. Kinds of Quantity. Homogeneous Equa- 
tions. Parallelogram of Forces 1 

PAET I. STATICS. 

Chaptek I. Statics of a Material Point. 
§§ 16-19. Composition and Equilibrium of Concurrent Forces. ... 8 

Chapter II. Parallel Forces and the Centre of Gravity. 

§§ 20-22. Parallel Forces 13 

§§ 23-276. Centre of Gravity. Problems. Centrobaric Method. . . 18 

Chapter III. Statics of a Rigid Body. 

§§28-34. Couples 27 

•§§ 35-39. Composition and Equilibrium of Non-concurrent Forces. 31 

Chapter IV. Statics of Flexible Cords. 

§§40-48. Postulates. Suspended Weights. Parabolic Cord. Cat- 
enary 42 

PAET II. DYNAMICS. 

Chapter I. Rectilinear Motion of a Material Point. 

§§ 49-55. Uniform Motion. Falling Bodies. Newton's Laws. 

Mass 49 

§§ 56-60. Uniformly Accelerated Motion. Graphic Representa- 

tions. Variably Accelerated Motions. Impact... 54 

Chapter II. Virtual Velocities. 

§§ 61-69, Definitions and Propositions. Connecting-rod. Prob- 
lems 67 

vii 



yiii CONTENTS. 

Chapter III. Curvilinear Motion op a Material Point. 

PAGB 

§§ 70-74. Composition of Motions, of Velocities, etc. General 

Equations 72 

§§ 75-84 Normal Acceleration. Centripetal and Centrifugal 
Forces. Simple Pendulums. Projectiles. Rela- 
tive Motion 77 

Chapter IV. Moment of Inertia. 

§§ 85-94. Plane Figures. Rigid Bodies, Reduction Formulae. 

The Rectangle, Triangle, etc. Compound Plane 

Figures. Polar Moment of Inertia 91 

§S 95-104. Rod. Thin Plates. Geometric Solids 9$ 

§§ 105-107, Numerical Substitution. Ellipsoid of Inertia 103 

Chapter V. Dynamics op a Rigid Body. 

§§ 108-115. Translation. Rotation about a Fixed Axis. Centre of 

Percussion ...... 105 

§§ 116-121. Torsion Balance. Compound Pendulum. The Fly- 
wheel 116 

§§122-123. Uniform Rotation. " Centrifugal Action." Free Axes. 125 

§§ 124-126. Rolling Motions. Parallel Rod of Locomotive 130 

Chapter VI. Work, Energy, and Power. 

§§ 127-134. Work. Power. Horse-power. Kinetic Energy 133 

§§135-138. Steam-hammer. Pile-driving. Inelastic Impact „ 138 

§§ 139-141. Rotary Motion. Equivalent Systems of Forces. Any 

Motion of a Rigid Body 143 

§§ 142-146. Work and Kinetic Energy in a Moving Machine of 

Rigid Parts 147 

§§ 147-155. Power of Motors. Potential Energy. Heat, etc. Dy- 
namometers. Atwood's Machine. Boat-rowing. 
Examples 153- 

Chapter VII. Friction. 

§§ 156-164. Sliding Friction. Its Laws. Bent Lever 164 

§§ 165-171. Axle - friction. Friction Wheels. Pivots. Belting. 

Transmission of Power by Belting 175 

§§ 172-177. Rolling Friction. Brakes. Engine-friction. Anoma- 
lies in Friction. Rigidity of Cordage. Examples. 18$ 



CONTENTS. 1X 

PAET III. STEENGTH OF MATEEIALS. 

(OR MECHANICS OF MATERIALS.) 

CHAPTER I. ELEMENTARY STRESSES AND STRAINS. 

§§ 178-182. Stress and Strain ; of Two Kinds. Oblique Section 

of Rod in Tension 195 

§§ 182a-190. Hooke's Law. Elasticity. Safe Limit. Elastic 
Limit. Rupture. Modulus of Elasticity. 
Isotropes. Resilience. Internal Stress. Tem- 
perature Stresses 201 

TENSION. 

§§ 191-195. Hooke's Law by Experiment. Strain Diagrams. 

Lateral Contraction. Modulus of Tenacity 207 

§§ 196-199. Resilience of Stretched Prism. Load Applied Sud- 
denly. Prism Under Its Own Weight. Solid 
of Uniform Strength. Temperature Stresses. . 213 

COMPRESSION OF SHORT BLOCKS. 

§§ 200-202. Short and Long Columns. Remarks on Crushing. . . 218 

EXAMPLES IN TENSION AND COMPRESSION. 

§§ 203-206. Tables. Examples. Factor of Safety. Practical 

Notes 220 

SHEARING. 

§§ 207-213. Rivets. Shearing Distortion. Table. Punching. 

Examples 225 

CHAPTER II. TORSION. 
§§ 214-220. Angle and Moment of Torsion. Torsional Strength, 

Stiffness, and Resilience. Non-Circular Shafts 233. 

§§ 221-223. Transmission of Power* Autographic Testing Ma- 
chine. Examples 238 

CHAPTER III. FLEXURE OF HOMOGENEOUS PRISMS UNDER 
PERPENDICULAR FORCES IN ONE PLANE. 

§§ 224-232a. The Common Theory. Elastic Forces. Neutral 
Axis. The " Shear" and "Moment." Flex- 
ural Strength and Stiffness. Radius of Curva- 
ture. Resilience 244 

ELASTIC CURVES. 

§§ 233-238. Single Central Load ; at Rest, and Applied Suddenly. 

Eccentric Load. Uniform Load. Cantilever. . 252; 



X CONTENTS. 

SAFE LOADS. 

§§ 239-246. Maximum Moment. Shear = x-Derivative of the 
Moment. Simple Beams With Various Loads. 
Comparative Strength and Stiffness of Rectan- 
gular Beams 262 

§§ 247-252. Moments of Inertia. " Shape Iron." I-Beams, Etc. 

Cantilevers. Tables. Numerical Examples. . . 273 

SHEARING STRESSES IN FLEXURE. 

§§ 253-257. Shearing Stress Parallel to Neutral Surface ; and in 
Cross Section. Web of I-Beam. Riveting of 
Built Beams 284 

SPECIAL PROBLEMS IN FLEXURE. 

§§ 258-265. Designing Sections of Built Beams. Moving Loads. 
Special Cases of Quiescent Loads. Hydrostatic 
Load 295 

§§ 266-270. Derivatives of Ordinate of Elastic Curve. Weight 

Falling on Beam. Crank Shaft. Other Shafts. 310 

CHAPTER IV. FLEXURE; CONTINUED. 

CONTINUOUS GIRDERS. 

§§ 271-276. Analytical Treatment of Symmetrical Cases of Beams 

on Three Supports ; also, Built in ; (see § 391.). . 320 

THE DANGEROUS SECTION IN NON-PRISMATIC BEAMS. 

§§ 277-279. Double Truncated Wedge, Pyramid, and Cone 332 

NON-PRISMATIC BEAMS OF UNIFORM STRENGTH. 

§§ 280-289. Parabolic and Wedge-Shaped Beams. I-Beam. 

Elliptical Beam and Cantilevers 335 

DEFLECTION OF BEAMS OF UNIFORM STRENGTH. 

§§ 290-293. Parabolic and Wedge-Shaped Beams. Special Prob- 
lems 342 



CHAPTER V. FLEXURE OF PRISMATIC BEAMS UNDER 
OBLIQUE FORCES. 

§§ 294-301. Gravity and Neutral Axes. Shear, Thrust, and 
Stress-Couple. Strength and Stiffness of Ob- 
lique Cantilever. Hooks. Crane 347 



CONTENTS. xi 

CHAPTER VI. FLEXURE OF LONG COLUMNS. 

§§303-308.' End-Conditions. Euler's Formula. "Incipient 
Flexure." Hodgkinson's Formulae. Rankine's 
Formula 363 

§§ 309-314. Radii of Gyration. Built Columns. Trussed Gird- 
ers. Buckling of Web-Plates. Examples 376 

CHAPTER VII. LINEAR ARCHES (OF BLOCKWORK.) 

§§ 315-323. Inverted Catenary. Parabolic Arch. Circular Arch. 

Transformed Catenary as Arch „ 386 

CHAPTER VIII. ELEMENTS OF GRAPHICAL STATICS 

§§ 324-326. Force Polygons. Concurrent and Non-Concurrent 
Forces in a Plane. Force Diagrams. Equili- 
brium Polygons 397 

§§ 327-332. Constructions for Resultant, Pier-Reactions, and 
Stresses in Roof Truss. Bow's Notation. The 
Special Equilibrium Polygon. 402 

CHAPTER IX. GRAPHICAL STATICS OF VERTICAL FORCES. 

§§ 333-336. Jointed Rods. Centre of Gravity 412 

§§ 337-343. Useful Relations Between Force Diagrams and Their 

Equilibrium Polygons. . . 415 

CHAPTER X. RIGHT ARCHES OF MASONRY. 

§§ 344-352. Definitions. Mortar and Friction. Pressure in Joints. 
Conditions of Safe Equilibrium. True Linear 

Arch. 421 

§§ 353-357. Arrangement of Data for Graphic Treatment 428 

§§ 358-362. Graphical Treatment of Arch. Symmetrical and 

Unsymmetrical Cases 431 

CHAPTER XI. ARCH-RIBS. 

§§ 364-374. Mode of Support. Special Equilibrium Polygon and 
its Force Diagram. Change in Angle Between 
Rib Tangents. Displacement of Any Point on 
Rib 438 

§§ 374o-378a. Graphical Arithmetic. Summation of Products. 
Moment of Inertia by Graphics. Classification 
of Arch-Ribs 450 

§§ 379-388. Prof. Eddy's Graphical Method for Arch-Ribs of 
Hinged Ends ; and of Fixed Ends. Stress 
Diagrams. Temperature Stresses. Braced 
Arches 461 



*U 



CONTENTS. 



HORIZONTAL STRAIGHT GIRDERS. 



§§ 389-390. Prismatic Girder Supported at the Extremities, also 

with Ends Fixed Horizontally 479 



CHAPTER XII. GRAPHICS OF CONTINUOUS GIRDERS. 



§§ 391-397. The Elastic Curve an Equilibrium Polygon. Mohr's 
Theorem. Example and Numerical Case. End- 
Tangents. Re-arrangement of Moment- Area. . 

§§ 398-405. Positive Moment- Areas ; Amount and Gravity Ver- 
tical. Construction of the "False Polygons" 
in Any Given Case. Moment Curves, Shears, 
and Reactions. Numerical Example in Detail. 
Remarks 



484 



497 



[CONTENTS OF "MECHANICS OF FLUIDS: 1 } 
PAKT IV.— HYDKAULICS. 

CHAPTER I.— DEFINITIONS. FLUID PRESSURE. HYDRO- 
STATICS BEGUK 

PAGE 

§§ 406-417. Perfect fluids. Liquids and Gases. Principle of ' ! Equal 

Transmission of Pressure." Non-planar Pistons .. . 515 

§§ 418-427. Hydraulic Press. Free Surface of Liquid. Barometers 
and Manometers. Safety-valves. Strength of Thin 
Hollow Cylinders against Bursting and Collapse . . . 526 

CHAPTER II.— HYDROSTATICS CONTINUED. PRESSURE OF 
LIQUIDS IN TANKS AND RESERVOIRS. 

§§428-434. Liquid in Motion, but in "Relative Equilibrium." 
Pressure on Bottom and Sides of Vessels. Centre 
of Pressure of Rectangles, Triangles, etc 540 

§§ 435-444. Stability of Rectangular and Trapezoidal Walls against 
Water Pressure. High Masonry Dams. Proposed 
Quaker Bridge Dam. Earthwork Dam. Water 
Pressure on both Sides of a Gate 554 



CHAPTER III.-EARTH PRESSURE AND RETAINING WALLS. 

§§ 445-455. Angle of Repose. Wedge of Maximum Thrust. Geo- 
metrical Constructions. Resistance of Retaining 
Walls. Results of Experience 572 



CHAPTER IV.— HYDROSTATICS CONTINTJED. 
AND FLOTATION. 



IMMERSION 



§§456-460. Buoyant Effort. Examples of Immersion. Specific 

Gravity. Equilibrium of Flotation. Hydrometer.. 586 

§§ 461-465. Depth of Flotation. Draught and Angular Stability of 

Ships. The Metacentre 592 

xiii 



XIV 



CONTENTS. 



CHAPTER V.— HYDROSTATICS CONTINUED. 
FLUIDS. 



GASEOUS 



§§466-478. Thermometers. Absolute Temperature. Gases and Va- 
pors. Critical Temperature. Law of Charles. 
Closed Air-manometer. Mariotte's Law. Mixture of 
Gases. Barometric Levelling. Adiabatic Change.. 604 

§§479-489. Work Done in Steam-engine Cylinders. Expanding 
Steam. Graphic Representation of Change of State 
of Gas. Compressed-air Engine. Air-compressor. 
Hot-air Engines. Gas-engines. Heat- efficiency. 
Duty of Pumping-engines. Buoyant Effort of the 
Atmosphere 624 

CHAPTER VI.— HYDRODYNAMICS BEGUN. STEADY FLOW 
OF LIQUIDS THROUGH PIPES AND ORIFICES. 

§§ 489a-495. Phenomena of a " Steady Flow." Bernoulli's Theorem 
for Steady Flow without Friction, and Applications. 
Orifice in " Thin Plate" 646 

§§ 496-500. Rounded Orifice. Various Problems involving Flow 
through Orifices. Jet from Force-pump. Velocity 
and Density; Relation. Efflux under Water. Efflux 
from Vessel in Motion. Barker's Mill 663 

§§ 501-508. Efflux from Rectangular and Triangular Orifices. Pon- 
celet's Experiments. Perfect and Complete Con- 
traction, etc. Overfall Weirs. Experiments of 
Francis, Fteley and Stearns, and Bazin. Short 
Pipes or Tubes 672 

§§509-513. Conical Tubes. Venturi's Tube. Fluid Friction. 
Froude's Experiments. Bernoulli's Theorem with 
Friction. Hydraulic Radius. Loss of Head. Prob- 
lems involving Friction Heads in Pipes. Accumu- 
lator 692 

§§ 513^-518. Loss of Head in Orifices and Short Pipes. Coefficient 
of Friction of Water in Pipes. Fanning's Table. 
Petroleum Pumping. Flow through Long Pipes. . 703 

§§ 519-526. Chezy's Formula. Fire-engine Hose. Pressure-energy. 
Losses of Head due to Sudden Enlargement of Sec- 
tion; Borda's Formula. Diaphragm in Pipe. Ven- 
turi Water-meter 714 

§§ 527-536. Sudden Diminution of Section. Losses of Head due to 
Elbows, Bends, Valve-gates, and Throttle-valves. 
Examples, Prof. Bellinger's Experiments on Elbows. 
Siphons. Branching Pipes. Time of Emptying 
Vessels of Various Forms; Prisms, Wedges Pyra- 
mids, Cones, Paraboloids, Spheres, Obelisks, and 
Volumes of Irregular Form using Simpson's Rule. . 727 



CONTENTS. XV 

PAGE 

CHAPTER VII.— HYDRODYNAMICS, CONTINUED; STEADY 
FLOW OF WATER IN OPEN CHANNELS. 

§§ 538-542a. Nomenclature. Velocity Measurements and Instru- 
ments for the same. Ritchie-Haskell Direction 
Current- meter. Change of Velocity with Depth. 
Pitot's Tube. Hydrometric Pendulum. Wolt- 
mann's Mill. Gauging Streams. Chezy's Formula 
for Uniform Motion in Open Channel. Experi- 
ments 749 

§§ 5425-547. Kutter's Formula. Sections of Least Resistance. Trape- 
zoidal Section of Given Side Slope and Minimum 
Friction. Variable Motion in Open Channel. 
Bends. Formula introducing Depths at End Sec- 
tions. Backwater 759 

CHAPTER VIIL— DYNAMICS OF GASEOUS FLUIDS. 

§§ 548-556. Theorem for Steady Flow of Gases without Friction. 
Flow through Orifices by Water-formula; with 
Isothermal Expansion ; with Adiabatic Expansion. 
Maximum Flow of Weight. Experimental Co- 
efficients for Orifices and Short Pipes. Flow con- 
sidering Velocity of Approach 773 

§§ 557-561^. Transmission of Compressed Air through Long Pipes. 
Experiments in St. Gothard Tunnel. Pipes of Vari- 
able Diameter. The Piping of Natural Gas 786 

CHAPTER IX.— IMPULSE AND RESISTANCE OF FLUIDS. 

§§ 562-569. Reaction of a Jet of Liquid. Impulse of Jet on Curved 
Vanes, Fixed and in Motion. Pitot's Tube. The 
California "Hurdy-gurdy." Impulse on Plates. 
Plates Moving in a Fluid. Plates in Currents of 
Fluid ... 798 

§§ 570-575. Wind-pressure. Smithsonian Scale. Mechanics of the 
Sail-boat. Resistance of Still Water to Immersed 
Solids in Motion. Spinning Ball, Deviation from 
Vertical Plane. Robinson's Cup anemometer. Re- 
sistance of Ships. Transporting Power of a Cur- 
rent. Fire-streams, Hose-friction, etc - 818 



MECHANICS OF ENGINEERING, 



PRELIMINAKY CHAPTEE. 

1. Mechanics treats of the mutual actions and relative mo- 
tions of material bodies, solid, liquid, and gaseous; and by 
Mechanics of Engineering is meant a presentment of those 

principles of pure mechanics, and their applications, which are 
of special service in engineering problems. 

2. Kinds of Quantity. — Mechanics involves the following 
fundamental kinds of quantity : Space, of one, two, or three 
dimensions, i.e., length, surface, or volume, respectively ; time, 
which needs no definition here; force and mass, as defined be- 
low; and abstract numbers, whose values are independent of 
arbitrary units, as, for example, a ratio. 

3. Force. — A force is one of a pair of equal, opposite, and 
simultaneous actions between two bodies, by which the state* 
of their motions is altered or a change of form in the bodies 
themselves is effected. Pressure, attraction, repulsion, and 
traction are instances in point. Muscular sensation conveys 
the idea of force, while a spring-balance gives an absolute 
measure of it, a beam-balance only a relative measure. In 
accordance with Newton's third law of motion, that action and 
reaction are equal, opposite, and simultaneous, forces always 
occur in pairs; thus, if a pressure of 40 lbs. exists between 
bodies A and B, if A is considered by itself (i.e., "free"), 
apart from all other bodies whose actions upon it are called 
forces, among these forces will be one of 40 lbs. directed from 
B toward A. Similarly, if B is under consideration, a force 

* The state of motion of a small body under the action of no force, or of 
balanced forces, is either absolute rest, or uniform motion in a right line. 
If the motion is different from this, the fact is due to the action of an un- 
balanced force (§ 54). 



2 MECHANICS OF ENGINEERING. 

of 40 lbs. directed from A toward B takes its place among the 
forces acting on B. This is the interpretation of Newton's 
third law. 

In conceiving of a force as applied at a certain point of a 
body it is useful to imagine one end of an imponderable spiral 
spring in a state of compression (or tension) as attached at the 
given point, the axis of the spring having the given direction 
of the force. 

4. Mass is the quantity of matter in a body. The masses of 
several bodies being proportional to their weights at the same 
locality on the earth's surface, in physics the weight is taken 
as the mass, but in practical engineering another mode is used 
for measuring it (as explained in a subsequent chapter), viz.: 
the mass of a body is equal to its weight divided by the ac- 
celeration of gravity in the locality where the weight is taken, 
or, symbolically, M— G -f- g. This quotient is a constant 
quantity, as it should be, since the mass of a body is invariable 
wherever the body be carried. 

5. Derived Quantities. — All kinds of quantity besides the 
fundamental ones just mentioned are compounds of the latter, 
formed by multiplication or division, such as velocity, accele- 
ration, momentum, work, energy, moment, power, and force- 
distribution. Some of these are merely names given for 
convenience to certain combinations of factors which come 
together not in dealing with first principles, but as a result of 
common algebraic transformations. 

6. Homogeneous Equations are those of such a form that they 

are true for any arbitrary system of units, and in which all 

terms combined by algebraic addition are of the same kind. 

at 2 
Thus, the equation s — ~ (in which g = the acceleration of 

gravity and t the time of vertical fall of a body in vacuo, 
from rest) will give the distance fallen through, s, whatever 
units be adopted for measuring time and distance. But if for 



PRELIMINARY CHAPTER. 



g we write the numerical value 32.2, which it assumes when 
time is measured in seconds and distance in feet, the equation 
$ = 16.1f is true for those units alone, and the equation is not 
of homogeneous form. Algebraic combination of homogeneous 
equations should always produce homogeneous equations ; if 
not, some error has been made in the algebraic work. If any 
equation derived or proposed for practical use is not homogene- 
ous, an explicit statement should be made in the context as to 
the proper units to be employed. 

7. Heaviness. — By heaviness of a substance is meant the 
weight of a cubic unit of the substance. E.g. the heaviness of 
fresh water is 62.5, in case the unit of force is the pound, 
and the foot the unit of space; i.e., a cubic foot of fresh 
water weighs 62.5 lbs. In case the substance is not uniform 
in composition, the heaviness varies from point to point. If 
the weight of a homogeneous body be denoted by G, its volume 
by V y and the heaviness of its substance by y, then G = Vy. 



Weight in Pounds of a Cubic Foot (i.e., the heaviness) of various 

Materials. 



Anthracite, solid 100 

" broken 57 

Brick, common hard 125 

" soft 100 

Brick-work, common 112 

Concrete 125 

Earth, loose 72 

as mud 102 

Granite 164 to 172 

Ice., 58 

Iron, cast 450 

" wrought 480 



Masonry, dry rubble 138 

" dressed granite or 

limestone 165 

Mortar 100 

Petroleum 55 

Snow ' 7 

" wet 15 to 50 

Steel 490 

Timber 25 to 60 

Water, fresh 62.5 

sea 64.0 



8. Specific Gravity is the ratio of the heaviness of a material 
to that of water, and is therefore an abstract number. 



9. A Material Point is a solid body, or small particle, whose 
dimensions are practically nothing, compared with its range of 
motion. 



4 MECHANICS OF ENGINEERING. 

10. A Rigid Body is a solid, whose distortion or change of 
form under any system of forces to be brought upon it in 
practice is, for certain purposes, insensible. 

11. Equilibrium. — When a system of forces applied to a 
body produces the same effect as if no force acted, so far as 
the state of motion of the body is concerned, they are said to 
be balanced, or to be in equilibrium. [If no force acts on a 
material point it remains at rest if already at rest ; but if 
already in motion it continues in motion, and uniformly 
(equal spaces in equal times), in a right line in direction 
of its original motion. See § 54.] 

12. Division of the Subject. — Statics will treat of bodies at 
rest, i.e., of balanced forces or equilibrium ; dynamics, of 
bodies in motion ; strength of materials will treat of the effect 
of forces in distorting bodies ; hydraulics, of the mechanics 
of liquids and gases (thus including pneumatics). 

13. Parallelogram of Forces. — Duchayla's Proof. To fully 
determine a force we must have given its amount, its direc- 
tion, and its point of application in the body. It is generally 
denoted in diagrams by an arrow. It is a matter of experience 
that besides the point of application already spoken of any 
other may be chosen in the line of action of the force. This 
is called the transmissibility of force; i.e., so far as the state of 
motion of the body is concerned, a force may be applied any- 
where in its line of action. 

The Resultant of two forces (called its components) applied 
at a' point of a body is a single force applied at the same point, 
which will replace them. To prove that this resultant is given 
in amount and position by the diagonal of the parallelogram 
formed on the two given forces (conceived as laid off to some 
scale, so many pounds to the inch, say), Duchayla's method 
requires four postulates, viz. : (1) the resultant of two forces 
must lie in the same plane with them ; (2) the resultant of two 
equal forces must bisect the angle between them ; (3) if one of 
the two forces be increased, the angle between the other force 
and the resultant will be greater than before; and (4) the trans- 
missibility of force, already mentioned. Granting these, we 
proceed as follows (Fig. 1) : Given the two forces P and Q = 



PRELIMINARY CHAPTER. 5 

jf* + P" {P' and P" being each equal to P, so that # = 2P), 
applied at 0. Transmit P" to A. Draw the parallelograms 
(?J5 and AD', OD will also be a parallelogram. By postulate 
(2), since OB is a rhombus, P and jP' at may be replaced by 
a single force R' acting through B. Transmit R' to B and 
replace it by P and P' . Transmit P from B to ^L, i 3 ' from 
B to P. Similarly P and .P", at ^4, may be replaced by a 
single force R" passing through D ; transmit it there and re- 
solve it into P and P" . P' is already at D. Hence P and 
P' -f- P'\ acting at D, are equivalent to P and P' -\- P" act- 
ing at 0, in their respective directions. Therefore the result- 
ant of P and P' -\- P" must lie in the line OD, the diagonal 
of the parallelogram formed on P and Q = 2P at 0. Similarly 



%J-9 




Fig. 1. 




this may be proved (that the diagonal gives the direction of 
the resultant) for any two forces P and mP\ and for any two 
forces nP and mP, m and n being any two whole numbers, 
i.e., for any two commensurable forces. When the forces are 
incommensurable (Fig. 2), P and Q being the given forces, 
we may use a reductio ad absurdum, thus : Form the parallelo- 
gram OP on P and Q applied at 0. Suppose for an instant 
that R the resultant of P and Q does not follow the diagonal 
OD, but some other direction, as OD' . Note the intersection 
II, and draw HC parallel to DB. Divide P into equal parts, 
each less than RD ; then in laying off parts equal to these from 
O along OB, a point of division will come at some point F 
between and B. Complete the parallelogram OFEG. The 
force Q" = OP is commensurable with P, and hence their 



b MECHANICS OF ENGINEERING. 

resultant acts along OK Now Q is greater than Q'\ while R 
makes a less angle with P than OE, which is contrary to pos- 
tulate (3); therefore R cannot lie outside of the line OD. 
Q.E.D. 

It still remains to prove that the resultant is represented in 

amount, as well as position, by the diagonal. OD (Fig. 3) is 

x/r' the direction of R the resultant of P and 

,/F \ Q; required its amount. If R' be a force 

"/if".; — ~>\ y equal and opposite to R it will balance P 

'\A \d/1 and Q ; i.e, the resultant of R' and P 
p £* must lie in the line QO prolonged (besides 

FlG ' 3- being equal to Q). We can therefore de- 

termine R' by drawing BA parallel to DO to intersect QO 
prolonged in A ; and then complete the parallelogram on 
BA and BO.* Since OFABis a parallelogram R' must =BA, 
and since OABD is a parallelogram BA=OD; therefore 
W^OD and also R=OD. Q. E. D. 

Corollary. — The resultant of three forces applied at the same 
point is the diagonal of the parallelo piped formed on the three 
forces. 

14. Concurrent forces are those whose lines of action intersect 
in a common point, while non-concurrent forces are those which 
do not so intersect ; results obtained for a system of concurrent 
forces are really derivable, as particular cases, from those per- 
taining to a system of non-concurrent forces. 

15. Resultant. — A single force, the action of which, as re- 
gards the state of motion of the body acted on, is equivalent to 
that of a number of forces forming a system, is said to be the 
Resultant of that system, and may replace the system ; and con- 
versely a force which is equal and opposite to the resultant of 
a system will balance that system, or, in other words, when it 
is combined with that system there will result a new system in 
equilibrium ; this (ideal) force is called the Anti-resultant. 

In general, as will be seen, a given system of forces can al- 

* R' must = ~OF; for if R' > or < OF, the diagonal formed on R' and P 
cannot take the direction of QO prolonged. 



PRELIMINARY CHAPTER. 7 

ways De replaced by two single forces, but these two can be 
combined into a single resultant only in particular cases. 

15a. Equivalent Systems are those which may be replaced by 
the same set of two single forces — or, in other words, those 
which have the same effect, as to state of motion, upon the 
given body. 

15b. Formulae. — If in Fig. 3 the forces P and Q and the angle a = 
PO Q are given, we have, for the resultant, 



R = D = <\/ P 2 + Q 2 + 2 P Q cos a. 

(If a is > 90° its cosine is negative.) In general, given any three parts 
of either plane triangle D Q, or D B, the other three may be obtained 
by ordinary trigonometry. Evidently if a = 0, P = P + Q ; if a = 
180°, P = P- Q; and if a = 90°, B = V P 2 + Q\ 

15c. Varieties of Forces. — Great care should be used in deciding 
what may properly be called forces. The latter may be divided into ac- 
tions by contact, and actions at a distance. If pressure exists between two 
bodies and they are perfectly smooth at the surface of contact, the pressure 
(or thrust, or compressive action), of one against the other constitutes a force, 
whose direction is normal to the tangent plane at any point of contact (a 
matter of experience) ; while if those surfaces are not smooth there may also 
exist mutual tangential actions or friction. (If the bodies really form a 
continuous substance at the surface considered, these tangential actions are 
called shearing forces.) Again, when a rod or wire is subjected to tension, 
any portion of it is said to exert a pull or tensile force upon the remainder ; 
the ability to do this depends on the property of cohesion. The foregoing 
are examples of actions by contact. 

Actions at a distance are exemplified in the mysterious attractions, or re- 
pulsions, observable in the phenomena of gravitation, electricity, and mag- 
netism, where the bodies concerned are not necessarily in contact. By the 
term weight we shall always mean the force of the earth's attraction on the 
body in question, and not the amount of matter in it. 

[Note.— In some common phrases, such as " The tremendous force "of a heavy body in 
rapid motion, the word force is not used in a technical sense, hut signifies energy (as ex- 
plained in Chap. VI.). The mere fact that a body is in motion, whatever its mass and 
velocity, does not imply that it is under the action of any force, necessarily. For instance, 
at any point in the path of a cannon ball through the air, the only forces acting on it are 
the resistance of the air and the attraction of the earth, the latter having a vertical and 
downward direction.] 



PART I -STATICS, 



CHAPTER I. 



STATICS OF A MATERIAL POINT. 




Fig. 4. 



16. Composition of Concurrent Forces. — A system of forces 
acting on a material point is necessarily composed of concurrent 
forces. 

Case I. — All the forces in One Plane. Let be the 
material point, the common point of application of all the 
forces ; P x , P„ etc., the given forces, making 
angles a v a„ etc., with the axis X. By the 
parallelogram of forces P. may be resolved 
into and replaced by its components, P x cos a x 
acting along X, and P x sin a x along Y. 
Similarly all the remaining forces may be re- 
placed by their X and Y components. We have now a new 
system, the equivalent of that first given, consisting of a set of 
X forces, having the same line of application (axis X), and a 
set of Y forces, all acting in the line Y. The resultant of the 
X forces being their algebraic sum (denoted by 2X) (since 
they have the same line of application) we have 

2X= P x cos a x -\- _P a cos <x a -J- etc. = ^(P cos a), 

and similarly 

2Y=P X sin a x -f- P 2 sin a 2 -f- etc. = 2(P sin a). 

These two forces, 2X and 2Y, may be combined by the 
parallelogram of forces, giving P = \/r^XY + (^Y) 2 as tne 
single resultant of the whole system, and its direction is deter- 

2Y 

mined by the angle **; thus, tan a = ^^; see Fig. 5. For 

equilibrium to exist, P must = 0, which requires, separately, 



STATICS OF A MATERIAL POINT. 9 

2X = 0, and 2Y = (for the two squares (2Xy and 
(2 Y) 2 can neither of them be negative quantities). 

Case II. — The forces having any directions in space, 
but all applied at 0, the material point. Let P x , P„ 
-etc., be the given forces, P x making the angles a^ j3 x , and y x , 
respectively, with three arbitrary axes, X, T", and Z (Fig. 6), 
at right angles to each other and intersecting at 0, the origin. 
Similarly let <* 2 , /? 2 , y„ be the angles made by P t with these 
.axes, and so on for all the forces. By the parallelopiped of 
forces, P x may be replaced by its components. 

X x = P x cos a 19 Y x = P x cos /?„ and Z x = P x cos y x ; and 






Fig. 5, 



Fig. 6. 



Fig. 7. 



-similarly for all the forces, so that the entire system is now 
replaced by the three forces, 

2X — P x cos a x -\- P 2 cos a 2 -f- etc ; 
2Y = P x cos fi x + P 2 cos /?„ -f etc; 
2Z = P x cos y x -\- P 2 cos y 2 -f etc ; 

and finally by the single resultant 

R = V(zxy + {2 ry + (szy. 

Therefore, for equilibrium we must have separately, 
2X= 0, 2Y= 0, and 2Z= 0. 
i?'s position may be determined by its direction cosines, viz., 

2X a 2Y 2Z 

— ; cos /3 = -g- ; cos y = -^-. 



cos a 



17. Conditions of Equilibrium. — Evidently, in dealing with 
a system of concurrent forces, it would be a simple matter to 



CO MECHANICS OF ENGINEERING. 

replace any two of the forces by their resultant (diagonal 
formed on them), then to combine this resultant with a third 
force, and so on until all the forces had been combined, the 
last resultant being the resultant of the whole system. The 
foregoing treatment, however, is useful in showing that for 
equilibrium of concurrent forces in a plane only two conditions 
are necessary, viz., 2X = and 2 Y = 0; while in space 
there are three, 2X = 0, 2Y = 0, and 2Z = 0. In Case L, 
then, we have conditions enough for determining two unknown 
quantities ; in Case II., three. 

18. Problems involving equilibrium of concurrent forces. 

(A rigid body in equilibrium under no more than three forces 

may be treated as a material point, since the (two or) three 

forces are necessarily concurrent.)* 

Problem 1. — A body weighing G lbs. rests on a horizontal 

table : required the pressure between it and the table. Fig. 8. 

Consider the body free, i.e., conceive all other bodies removed 
, (the table in this instance), being replaced by the 

j forces which they exert on the first body. Taking 

|G the axis T' vertical and positive upward, and not 



Of- 



+X assuming in advance either the amount or direc- 
ts tion of X, the pressure of the table against the 
body, but knowing that G, the action of the earth 

fig. 8. on ^ ie body, is vertical and downward, we have 
here a system of concurrent forces in equilibrium, in which 
the X and Y components of G are known (being and — 
G respectively), while those, X x and X Y , of X are unknown. 
Putting 2X = 0, we have X x — [- 0=0; i.e., JVhas no hori- 
zontal component, .*. X is vertical. Putting 2 Y = 0, we 
have X Y — G = 0, .*. X Y = -f- G ; or the vertical component 
of X, i.e., X itself, is positive (upward in this case), and is 
numerically equal to G. 

Problem 2. — Fig. 9. A body of weight G (lbs.) is moving 
in a straight line over a rough horizontal table with a uniform 
velocity v (feet per second) to the right. The tension in an 
oblique cord by which it is pulled is given, and = P (lbs.), 

* Three parallel forces form an exception ; see §§ 20, 21, etc. 



STATICS OF A MATERIAL POINT. 11 

which remains constant, the cord making a given angle of 
elevation, a, with the path of the body. Required the vertical 
pressure N (lbs.) of the table, and also its + y 

horizontal action F (friction) (lbs.) against p 

the body. „ ,-* — rf\ 

Referring: by anticipation to Newton's first ,"1"""""""' " " 
law of motion, viz., a material point acted *^WMMfy\4/ 
on by no force or by balanced forces is either FlG - 9. 

at rest or moving uniformly in a straight line, we see that this 
problem is a case of balanced forces, i.e., of equilibrium. Since 
there are only two unknown quantities, N and F, we may 
determine them by the two equations of Case I., taking the 
axes X and Y as before. Here let us leave the direction of 
i^as well as its amount to be determined by the analysis. As 
T^miist evidently point toward the left, treat it as negative in 
summing the X components ; the analysis, therefore, can be 
expected to give only its numerical value. 
2X = gives P cos oc — F= 0. .*. F = P cos a. 

2Y=0givesJST+Psma- G = 0. .\ JV = G - P sin a. 
.'. JV is upward or downward according as G is > or < P 
sin a. For W to be a downward pressure upon the body would 
require the surface of the table to be above it. The ratio of the 
friction F to the pressure JX which produces it can now be 
obtained, and is called the coefficient of friction. It may vary 
slightly with the velocity. 

This problem may be looked upon as arising from an experi- 
ment made to determine the coefficient of friction between the 
given surfaces at the given uniform velocity. 

19. The Free-Body Method. — The foregoing rather labored so- 
lutions of very simple problems have been made such to illus- 
trate what may be called the free-body method of treating any 
problem involving a body acted on by a system of forces. It 
consists in conceiving the body isolated from all others which 
act on it in any way, those actions being introduced as so many 
forces, known or unknown, in amount and position. The sys- 
tem of forces thus formed may be made to yield certain equa- 



12 MECHANICS OF ENGINEERING. 

tions, whose character and number depend on circumstances, 
such as the behavior of the body, whether the forces are con- 
fined to a plane or not, etc., and which are therefore theoreti- 
cally available for determining an equal number of unknown 
quantities, whether these be forces, masses, spaces, times, or 
abstract numbers. Of course in some instances the unknown 
quantities may enter these equations with such high powers 
that the elimination may be impossible ; but this is a matter 
of algebra, not of mechanics. 

(Addendum to § 49a of page 49.) Numerical Example.— A set of light 
■screens is set up at intervals of 100 feet apart in the horizontal path of a 
cannon-ball, with the object of determining its velocity, and also the rate of 
•change (or negative acceleration) of that velocity, as due to the resistance 
of the air. 

By electrical connection the time of passing each screen is noted, and 
the intervals of time are given in this diagram for four of the screens, 
A, B, C, and D. 



100' 



..0.0621 sec. 



....100' 

.0.0632 sec... 



.100' 



..0.0643 sec. 



A 1 B 2 G 3 D 

From these data it is required to compute, as nearly as the circumstances 
allow, the velocity and acceleration (negative) of the ball at various points 
(the ball moves from left to right). 

Solution. — In passing from A to B the ball has an average velocity of 
1610 ft. per second, obtained by dividing the distance of 100 feet by the 
time of passage, 0.0621 second. Similarly we find the average velocity 
between B and G to be 1582 ft. per second, and that between G and B to 
be 1554 ft. per second. 

As the velocity is not changing very rapidly, we may claim that the ball 
actually possesses the velocity v x = 1610 ft. per second at the point 1, mid- 
way between A and B, or very near that point ; and similarly the velocity 
v 2 = 1582 ft. per second at point 2, midway between B and C ; and 
v 3 = 1554 ft. per second at point 3, midway between C and D. 

Hence the total gain of velocity from 1 to 2 is 1582-1610 = — 28 ft. per 
second; and the time in which this gain is made is one half of the 0.0621 
second plus one half of the 0.0632 second, i.e., 0.0626 second. Therefore 
an approximate value for the average acceleration between points 1 and 2 
is found by dividing the — 28 ft. per second gain in velocity by the 
time 0.0626 second occupied in acquiring the gain. This gives — 447 ft. 
per second per second average acceleration for portion 1. , .2 of path, and 
since screen B lies at the middle of this portion, the actual acceleration of 
the ball's motion as it passes the screen B is very nearly equal to this, viz.: 
— 447 ft. per second per second (or " ft. per square second "). 

By a similar process the student may compute the acceleration at screen 
G. Of course the reason why these results are merely approximate is that 
the spaces and times concerned, though relatively small, are not infinitesimal. 

[A recent English writer calls a unit of velocity a "speed ;" and a unit 
of acceleration, a "hurry."'] 



PARALLEL FORCES AND THE CENTRE OF GRAVITY. 13. 



CHAPTER II. 



PARALLEL FORCES AND THE CENTRE OF GRAVITY. 



20. Preliminary Remarks. — Although by its title this section 
should be restricted to a treatment of the equilibrium of forces, 
certain propositions involving the composition and resolution 
of forces, without reference to the behavior of the body under 
their action, will be found necessary as preliminary to the prin- 
cipal object in view. 

As a rigid body possesses extension in three dimensions, to 
deal with a system of forces acting on it we require three co- 
ordinate axes : in other words, the system consists of " forces 
in space," and in general the forces are non-concurrent. In 
most problems in statics, however, the forces acting are in one 
plane: we accordingly begin by considering non-concurrent 
forces in a plane, of which the simplest case is that of two 
parallel forces. For the present the body on which the forces 
act will not be shown in the figure, but must be understood to 
be there (since we have no conception of forces independently 
of material bodies). The device will frequently be adopted of 
introducing into the given system two opposite and equal forces 
acting in the same 4ine : evidently this will not alter the effect 
of the given system, as regards the rest or motion of the body. 

p'r-- TP 

21. Resultant of two Parallel 
Forces. I \ 

•p In l____\ 

Case I. — The two forces have ir- 
the same direction. Fig. 10. 
Let P and Q be the given forces, 
and AB a line perpendicular to 
them (P and Q are supposed to have sL 

been transferred to the intersections FlG - 10 - 

A and B). Put in at A and B two equal and opposite 
forces 8 and S, combining them with P and Q to form P' 



■ \-a — 

■x-A 



Q 
i ■ 



A\ 



/B S. 



cV ; ' 



m 



14 



MECHANICS OF ENGINEERING. 



have" 5 



Now write this 



and Q ' . Transfer P' and Q' to their intersection at C, and there 
resolve them again into S and P, S and Q. S and S annul each 
other at G\ therefore P and Q, acting along a common line CD, 
replace the P and Q first given ; i.e., the resultant of the origi- 
nal two forces is a force R =P -\- Q, acting parallel to them 
through the point D, whose position must now be determined. 
The triangle CAD is similar to the triangle shaded by lines, 
.*. P : S :: CD : x; and CDB being similar to the triangle 
shaded by dots, .*. S. : Q :: a — x : CD. Combining these, we 

P^Zl and .-. »=■-&-=& 

<?~ x P+Q R' 

Rx = Qa, and add Re, i.e., Pc -\- Qc, to each member, c being 
the distance of (Fig. 10), any point in AB produced, from 
A. This will give R(x -f- c) = Pc -f$( a + c )> i ]1 which c, 
<i'-\- c, and x -f- c are respectively the lengths of perpendiculars 
let fall from upon P, Q, and their resultant R. Any one of 
these products, such &sPc< is for convenience (since products of 
this form, occur so frequently in Mechanics as a result of alge- 
braic transformation) called the Moment of the force about the 
arbitrary point 0. Hence the resultant of two parallel forces of 
the same direction is equal to their sum, acts in their plane, in 
a line parallel to them, and at such a distance from any arbi- 
trary point in their plane as may be determined by writing 
its moment about equal to the sum of the moments of the 
two forces about 0. is called a centre of moments, an d each 
of the perpendiculars a lever-arm. 

Case II. — Two parallel forces P and Q of opposite direc- 
tions. Fig. 11. By a process similar to the foregoing, we 

obtain R =P- <)and \P - Q)x 
= Qa, i.e., Rx = Qa. Subtract 
each member of the last equation 
from Re (i.e., Pc—Qc), in which c 
is the distance, from A, of any arbi- 
trary point in AB produced. This 
gives R(c — x) = Pc — Q(a -f- c). 
But (c — a?), c, and (a -j- c) are re- 
spectively the perpendiculars, from 

* That is, the resultant of two parallel forces pointing in the same direc- 
tion divides the distance bet'uoeen them in the inverse ratio of those forces. 



1 i 


"P 






-k— -\ 


Co. 






Qj 


ft> 






D 


si„J 




\B 


k e-H 

i 


- -4A 

*-;t-~H 
f- — 


— -a-— 


Of 

— a 



PARALLEL FORCES AND THE CENTRE OF GRAVITY. 1 5 

O, upon R, P, and Q. • That is, R(e — x) is the moment of R 
about 0\ Re, that of R about 0; and Q{a-\-c), that of Q 
about 0. But the moment of Q is subtracted from that of R, 
which corresponds with the fact that Q in this figure would 
produce a rotation about opposite in direction to that of P. 
Having in view, then, this imaginary rotation, we may define 
the moment of a force as positive when the indicated direction 
about the given point is against the hands of a watch; as nega- 
tive when with the hands of a watch.* 

Hence, in general, the resultant of any two parallel forces is, 
in amount, equal to their algebraic sum, acts in a parallel direc- 
tion in the same plane, while its moment, about any arbitrary 
point in the plane, is equal to the algebraic sum of the mo- 
ments of the two forces about the same point. 

Corollary. — If each term in the preceding moment equations 
be multiplied by the secant of an angle (or, Fig. 12) thus: 



Pf 



Of---'- — ■&] — J 






Xt 



JL 



a — > i 

Fig. 12. 




Fig. 13. 



(using the notation of Fig. 12), we have Pa sec a = P x a x 
sec a -J- P 2 <z 2 sec <*, i.e., Ph = Pfi l -j- P 2 b 2 , in which b, h x , 
and 5 2 are the oblique distances of the three lines of action 
from any point in their plane, and lie on the same straight 
line; P is the resultant of the parallel forces P x and P v 



22. Resultant of any System of Parallel Forces in Space. — 
LetP^Pv _P 3 , etc., be the forces of the system, and a? 15 y„ 
s„ a? 2 , 2/ 2 , s 2 , etc., the co-ordinates of their points of application 
as referred to an arbitrary set of three co-ordinate axes X, Y y 
and Z, perpendicular to each other. Each force is here re 

* These two directions of rotation are often called counter clockwise, and 
clockwise, respectively. 



16 MECHANICS OF ENGINEERING. 

suicted to a definite point of application in its line of action) 
(with reference to establishing more directly the fundamental 
equations for the co-ordinates of the centre of gravity of a. 
body). The resultant P' of any two of the forces, as 
P\ aud P„ is = P l -\- P„ and may be applied at (7, the in- 
tersection of its own line of action with a line BD joining 
the points of application of P 1 and P„ its components. 
Produce the latter line to A, where it pierces the plane XY y 
and let b v b', and 5 a , respectively, be the distances of B, O y 
P, from A • then from the corollary of the last article we have 

Pl' = P& + PJ>,; 

but from similar triangles 

V ibr-hnz':^: z„ .'. P'z' = P& + P,z v 

Now combine P', applied at (7, with P„ applied at E, calling* 
their resultant P" and its vertical co-ordinate z" , and we obtain 

P"z" = P'z' + P 3 z„ i.e., P"z" = P& + P A + P 3% 

also 

2" = P> + P 3 = P x + P,+ p s . 

Proceeding thus until all the forces have been considered, we- 
shall have finall y, for the resultant of the whole system, 

22 = P, + P„+P, + eta; 

and for the vertical co-ordinate of its point of application,, 
which we may /frite s, 

. - P lgl + P,g, + P,g,... _ 2(Pg) t 

i.e.,a j? p + p 2+ p 3 ^__ - 2 P ' 

and similarly (?rthe other co-ordinates. 

2(Px) - 2(Py) 

* = -gp- and y = sp-- 

In these equations, in the general case, such products as P x z lY 
etc., cannot strictly be called moments. The point whose co- 



PARALLEL FORCES AND THE CENTRE OF GRAVITY. 17 

oidinates are the a?, y, and s, just obtained, is called the Centre 
of Parallel Forces, and its position is independent of the {com- 
mon) direction of the forces concerned. 

Example. — If the parallel forces are contained in one plane, 
and the axis l^be assumed parallel to the direction of the 
forces, then each product like Pjx x will be a moment, as de- 
fined in § 21 ; and it will be noticed in the accompanying nu- 
merical example, Fig. 14, that a detailed substitution in the 
equation R D |Y R p 

■-% — 3 . — 



Ex = P,x t + P,x 2 + etc., ... (1) 
having regard to the proper sign of each 



t 



Ol +x 

force and of each abscissa, gives the same FlG . 14 . 

result as if each product Px were first obtained numerically, 
and a sign affixed to the product considered as a moment 
about the point 0. Let P x — — 1 lb.; P 2 = + 2 lbs.; P 3 = 
+ 3 lbs.; P 4 = - 6 lbs.; x, = +l ft.; x 2 = + 3 ft.; x a = - 2 
ft.; and a? 4 = — 1 ft. Required the amount and position of the 
resultant R. In amount E = 2P = — 1 + 2 + 3 — 6 = — 2 
lbs.; i.e., it is a downward force of 2 lbs. As to its position, 
Bx= 2{Px) gives ( - 2)aJ = (- 1) X (+ 1) + 2 X 3 + 
3 x (- 2) + (-6) X (- 1) = - 1 + 6 - 6 + 6. Now from 
the figure, by inspection, it is evident that the moment of P x 
about is negative {with the hands of a watch), and is numer- 
ically = 1, i.e., its moment == — 1 ; similarly, by inspection, 
that of P 2 is seen to be positive, that of P z negative, that of 
P i positive; which agree with the results just found, that 
(- 2)5 = — 1 + 6 — 6 + 6 = + 5 ft. lbs. (Since a moment 
is a product of a force (lbs.) by a length (ft.), it may be called 
so many foot-pounds.) Next, solving for x, we obtain 
x = (+ 5) -T- (^- 2) = — £5 ft.; i.e., the resultant of the given 
forces is a downward force of 2 lbs. acting in a vertical line 
2.5 ft. to the left of the origin. Hence, if the body in question 
be a horizontal rod whose weight has been already included in 
the statement of forces, a support placed 2.5 ft. to the left of 
and capable of resisting at least 2 lbs. downward pressure 
will preserve equilibrium ; and the pressure which it exerts 



18 MECHANICS OF ENGINEERING. 

against the rod must be an upward force, P b , of 2 lbs., i.e. ^e 
equal and opposite of the resultant of jP i5 P„ P % , P a . 

Fig. 15 shows the rod as a free body in equilibrium under 
the five forces. P b = -f- 2 lbs. = the reaction of the support. 

Of course P 5 is one of a pair of equal 

and opposite forces ; the other one 

■j is the pressure of the rod against the 



f 



P ?2 

A 



P 6 1 — 2:5- -io support, and would take its place among 

fig. 15. the forces acting on the support. 

23. Centre of Gravity. — Among the forces acting on any 
rigid body at the surface of the earth is the so-called attraction 
of the latter (i.e., gravitation), as shown by a spring-balance, 
which indicates the weight of the body hung upon it. The 
weights of the different particles of any rigid body constitute a 
system of parallel forces (practically so, though actually slightly 
convergent). The point of application of the resultant of these 
forces is called the centre of gravity of the body, and may also 
be considered the centre of mass, the body being of very small 
dimensions compared with the earth's radius. 

If x, y, and z denote the co-ordinates of the centre of gravity 
of a body referred to three co-ordinate axes, the equations 
derived for them in § 22 are directly applicable, with slight 
changes in notation. 

Denote the weight of any particle of the body by dG, its 
volume by dV, by y its heaviness (rate of weight, see § 7) and 
its co-ordinates by a?, y, and z ; then, using the integral sign as 
indicating a summation of like terms for all the particles of the 
body, we have, for heterogeneous bodies, 

•~_ fyxdV -_ fyydV m - _ fyzd V^ 

fydV V fydV" ~ fydV ' ' K) 
while, if the body is homogeneous, y is the same for all its ele- 
ments, and being therefore placed outside the sign of summa- 
tion, is cancelled out, leaving for homogeneous bodies ( V de- 
noting the total volume) 

- fxdV - fydV - fzdV 

x= J -y-\ y = — T — ; andz^* 7 -^.. . (2) 



PARALLEL FOECES AND THE CENTRE OF GRAVITY. 19 

Corollary. — It is also evident that if a homogeneous body is 
for convenience considered as made up of several finite parts, 
whose volumes are V 19 V„ etc., and whose gravity co-ordinates 
are x iy y» z x ; a? 2 , y„ 2, ; etc., we may write 

If the body is heterogeneous, put O x (weights), etc., instead 
of V„ etc., in equation (3). 

If the body is an infinitely thin homogeneous shell of uni- 
form thickness = A, then dV = hdF \dF denoting an element, 
and 7^ the whole area of one surface) and equations (2) become, 
after cancellation, 

^-f^l. z-fy^, -,-J^E ( 4 ) 

Similarly, for a homogeneous wire of constant small cross- 
section (i.e.. a geometrical line, having weight), its length 
being s, and an element of length ds, we obtain 

^ = Mi, yjyia.- z= f^ . . . (5) 

s s s 

It is often convenient to find the centre of gravity of a thin 
plate by experiment, balancing it on a needle-point; other 
shapes are not so easily dealt with. 

24. Symmetry. — Considerations of symmetry of form often 
determine the centre of gravity of homogeneous solids without 
analysis, or limit it to a certain line or plane. Thus the centre 
of gravity of a sphere, or any regular polyedron, is at its centre 
of figure; of a right cylinder, in the middle of its axis; of a 
thin plate of the form of a circle or regular polygon, in the 
centre of figure ; of a straight wire of uniform cross-section, in 
the middle of its length. 

Again, if a homogeneous body is symmetrical about a plane, 
the centre of gravity must lie in that plane, called a plane of 



jY /jAr 

^<;<:::-'-""' : f 4 j j 


TM 


>K-- a: -hi 

\. dxi 


-*— 



20 MECHANICS OF ENGINEERING. 

gravity ; if about a line, in that line called a line of gravity ; 
if about a point, in that point. 

25. By considering certain modes of subdivision of a homo- 
geneous body, lines or planes of gravity are often made appar- 
ent. E.g., a line joining the middle of the bases of a trape- 
zoidal plate is a line of gravity, since it bisects all the strips 
of uniform width determined by drawing parallels to the 
bases ; similarly, a line joining the apex of a triangular plate to 
the middle of the opposite side is a line of gravity. Other 
cases can easily be suggested by the student. 

26. Problems. — (1) Required the position of the centre of 
A gravity of & fine homogeneous wire of the 

-^ form of a circular arc, AB, Fig. 16. Take 
the origin at the centre of the circle, and 
the axis X bisecting the wire. Let the 
length of the wire, s, — 2s x ; ds = ele- 
ment of arc. We need determine only the 
*§•!// x, since evidently y = 0. Equations (5), 

fig. 16. g 23, are applicable here, i.e., x = . 

From similar triangles we have 

rdy 
ds-.dyv.r: x; .*. ds — —; 

:.~x — ^- Cdv — 7T— , i.e., = chord X radius -^- length of 

wire. For a semicircular wire, this reduces to x = %r -f- tc. 

Problem 2. Centre of gravity of trapezoidal {and trian- 
gular) thin plates, homogeneous, etc. — Prolong the non-parallel 
sides of the trapezoid to intersect at 0, which take as an origin, 
making the axis X perpendicular to the bases b and b x . We 
may here use equations (4), § 23, and may take a vertical strip 
for our element of area, dF, in determining x ; for each point 
of such a strip has the same x. Now dF = (y + y')dx. and 



PARALLEL FORCES AND THE CENTRE OF GRAVITY. 21 



from similar triangles y -f- y' = j x. Nowi^, = -(bh — bji^ 
can be written ^ , (A 2 — h*), and a? = -^ — becomes 



1 b 



UiJh, J 2 h K n *> 3 A 2 - A, 3 

for the trapezoid. 



For a triangle A x = 0, and we have 



A ; that is, the 



centre of gravity of a triangle is one third the altitude from the 
base. The centre of gravity is finally determined by knowing 



, 












y 


h 


( i^----^. 


-x-> 


V' 





*^k 




Fig. 17. 



Fig. 18. 



that a line joining the middles of b and b x is a line of gravity; 
or joining and the middle of b in the case of a triangle. 

Problem 3. Sector of a circle. Thin plate, etc. — Let the 
notation, axes, etc., be as in Fig. 18. Angle of sector = 2a ; 
x = % Using polar co-ordinates, the element of area dF (a 
small rectangle) == pdq> . dp, and its x = p cos <p ; hence the 
total area = 

i.e., F = r 2 a. From equations (4), § 23, we have 
- 1 /» 

= jv J cos <pp*dpd<p = yj a [_ GOS( pJ p*dp\ &<p- 



22 MECHANICS OP ENGINEERING. 

(Note on double integration. — The quantity 
cos cp J p 2 dp \dtp, 

is that portion of the summation / / cos cpp*dpdcp which 

belongs to a single elementary sector (triangle), since all its 
elements (rectangles), from centre to circumference, have the 
same cp and dcp.) 
That is, 

— 1 r 3 /* + * r 3 r+ a 2 r sin a 

* i i - "4 r sin -J- /? 
or, putting p = 2a = total angle of sector, a? = ^ -5 — . 

— 4?* 

For a semicircular plate this reduces to x = -^— . 

[Note. — In numerical substitution the arcs a and ft used 
above (unless sin or cos is prefixed) are understood to be ex- 
pressed in circular measure (^"-measure) ; e.g., for a quad- 



7t _ ^__ . „ _ 7C 



rant, p = ^ = 1*5707 + ; for 30°, p = g- ; or, in general, if /? 

. -, 180 ° 1 ^ 1 

in degrees == , then p in ^--measure = — . 

& n n _\ 

/\\ Problem 4. Sector of a flat ring ; thin 

_ V J° plate, etc. — Treatment similar to that of 

t Problem 3, the difference being that the 

t— pi 

/ limits of the interior integrations are 

// P 

/ instead of . Kesult, 

I — n 



Fig. 19 



- _ 4 y^ — r 3 sin -|/9 

a - 3 • r > _ v f » • ~ p ■■ 



PARALLEL FORCES AND THE CENTRE OF GRAVITY. 23 

Problem 5. — Segment of a circle ; thin plate, etc 
Since each rectangular element of any ver- 
tical strip has the same x, we may take the 
strip as dF 'in finding x, and use y as the 
half-height of the strip. dF = %ydx, and 
from similar triangles x : y : : ( — dy) : dx, 
i.e., xdx 
§23, 



— ydy. Hence from eq. (4), 




x = 



fxdF 
F 



Jx2ydx 
~~F^~~ 



- tftfdy 



_2_ 
SF 






Fig. 20. 



y 3 ) 



2 <? m 
d'F" 



but a = the half-chord, hence, finally, a? = 9 , / -. 

Problem 6. — Trapezoid; thin plate, etc., 
by the method in the corollary of § 23 ; equa- 
tions (3). Required the distance x from the 
base AB. Join BB, thus dividing the trape- 
zoid ABCD into two triangles ABB = F x 
and BBC = F 2 , whose gravity a?'s are, re- 
spectively, x x = -JA and a? 2 — $A. Also, F x 
= ihh^ F 2 = ihb„ and F (area of trape- 
zoid) = ihfa + &,). Eq. (3) of § 23 gives 
Fx_= F x x x -f- F& ; hence, substituting,^ -f- 
J b,)x=io 1 h+%b i h. 

Fio. 21. 




-_h (b, + 25,) 

"f-a: j, + j, ■ 

The line joining the middles of b x and & 2 is a line of gravity, and 
is divided in such a ratio by the centre of gravity that the fol- 
lowing construction for finding the latter holds good : Prolong 
each base, in opposite directions, an amount equal to the other 
base; join the two points thus found: the intersection with 
the other line of gravity is the centre of gravity of the trape- 
zoid. Thus, Fig. 21, with BF = b t and BF= b l9 join FF, 
etc. 



24 MECHANICS OF ENGINE K KING. 

Problem 7. Homogeneous oblique cone' or pyramid. — 
Take the origin at the vertex, and the axis X perpendicular to 
the base (or bases, if a frustum). In finding x we may put 
dV = volume of any lamina parallel to YZ, F being the base 
of such a lamina, each point of the lamina having the same x. 
Hence, (equations (2), § 23), 



but 



and 



x = ^fxdV, V=fdV=fFdx; 



F:F,::x>:h;, r.F = fix% 



F-S.2-, F* 



~i> fxdr= P x ° dx= ^Ii- 



— 3 h 4 — h 4 

For a frustum, x — —. ~ y^; while for a pyramid, h lt be- 

tfc n<2 — /&J 

— 3 
ing = 0, x = jA, Hence the centre of gravity of a pyramid 



is one fourth the altitude from the base. It also lies in the line 
h T - --7+\ joining the vertex to the centre of gravity 



z yf ■ 

(p! ) of the base. 



F) _>K Problem 8. — If the heaviness of the ma- 

&^\ [—, terial of the above cone or pyramid varied 

__// directly as x, y^ being its heaviness at the 

fig. 22. Dase F^ we wo aid use equations (1), § 23, 

y 

putting y = j- x ; and finally, for the frustum, 



4 h: - \ 

x -$ ' A 4 -A, 



- 4 

and for a complete cone x = -? A a . 



4 5 



27. The Centrobaric Method. — If an elementary area dF be 
revolved about an axis in its plane, through an angle a < 27r. 



PARALLEL FORCES AND THE CENTRE OF GRAVITY. 25 




the distance from the axis being == x, the volume generated is 
dV = axdF, and the total volume generated bj all the dF's 
of a finite plane figure whose plane con- 
tains the axis and which lies entirely on one 
side of the axis, will be V = fdV = 
afxdF But from §23, afxdF = cxFx; 
ax being the length of path described by 
the centre of gravity of the plane figure, FlG . * 

we may write : The volume of a solid of revolution generated 
by a plane figure, lying on one side of the axis, equals the 
area of the figure multiplied by the length of curve described 
by the centre of gravity of the figure. 

A corresponding statement may be made for the surface 
generated by the revolution of a line. The arc a must be ex- 
pressed in n measure in numerical work. 

27a. Centre of Gravity of any Quadrilateral. — Fig. 23$. 
B Construction; ABGD being any quad- 
rilateral. Draw the diagonals. On the 
long segment DK of DB lay off DE — 
BK, the shorter, to determine E; simi- 
larly, determine N on the other diagonal, 
by making GN = AK. Bisect EK in II 
and KN in M. The intersection of EM 
and NH \s the centre of gravity, O. 
Proof.— H being the middle of DB, and AH and HG 
having been joined, I the centre of gravity of the triangle 
ABD is found on AH, by making HI = \AH; similarly, by 
making HL = \HG, L is the centre of gravity of triangle 
BDG. . * . IL is parallel to AG and is a gravity-line of the 
whole figure ; and the centre of gravity C may be found on it 
if we can make CL : CI :: area ABD : area BDG (§ 21). 
But since these triangles have a common base DB, their areas 
are proportional to the slant heights (equally inclined to DB) 
AK and KG, i.e., to GN and NA. Hence HN, which di- 
vides IL in the required ratio, contains C, and is .'. a gravity- 
line. By similar reasoning, using the other diagonal, AG, and 




Fig. 23a. 



26 MECHANICS OF ENGINEERING. 

the two triangles into which it divides the whole figure, we 
may prove EM to be a gravity- line also. Hence the construc- 
tion is proved. 

27b. Examples. — 1. Required the volume of a sphere by 
the centrobaric method. 

A sphere may be generated by a semicircle revolving about 

its diameter through an arc a = 2n. The length of the path 

4-r 
described by its centre of gravity is = 27T^— (see Prob. 3, § 

26), while the area of the semicircle is inr" 2 . Hence by § 27, 

4:T 4 

Yolume generated = %n . — . \nr* = -g 7tr\ 

2. Required the position of the centre of gravity of the sector 
of a flat ring in which r x — 21 feet, r^ = 20 feet, and fi = 80° 
(see Fig. 19, and § 26, Prob. 4). 

P 

sin — == sin 40° = 0.64279, and fi in circular measure = 

-—- n — — it = 1.3962. By using r x and r a in feet, x will be 



= 18.87 feet. 



obtained in feet. 






_ 4 r 3 — t 3 
3 r'— r. 


sing 


4 1261 0.64279 
_ 3 ' 41 " 1.3962 



STATICS OF A KIGID BODY. 27 



CHAPTER III. 

STATICS OF A RIGID BODY. 

28. Couples. — On account of the peculiar properties and 
utility of a system of two equal forces acting in parallel lines 
and in opposite directions, it is specially /^? 
considered, and called a Couple. The ^^^ X 
arm of a couple is the perpendicular ^t^\sr? .--<•? 
distance between the forces ; its moment, \ <{a |oJ<^ j ^> 
the product of this arm, by one of the <^j ^^^ Ji^^ 
forces. The axis of a couple is an ^^^^^^^ 
imaginary line drawn perpendicular to FlG . 2 4. 

its plane on that side from which the rotation appears positive- 
(against the hands of a watch). (An ideal rotation is meant, 
suggested by the position of the arrows ; any actual rotation, 
of the rigid body is a subject for future consideration.) In 
dealing with two or more couples the lengths of their axes are 
made proportional to their moments; in fact, by selecting a 
proper scale, numerically equal to these moments. E.g., in Fig. 
24, the moments of the two couples there shown are Pa and 
Qb\ their axes p and q so laid off that Pa : Qb :: p : q, and 
that the ideal rotation may appear positive, viewed from the 
outer end of the axis. 

29. No single force can balance a couple. — For suppose the 
couple P, P, could be balanced by a force i?', then this, acting 

?t at some point C, ought to hold the couple 

pi /:::^_ B C. in equilibrium. Draw CO through 0, the 

*T fp $* centre of symmetry of the couple, and 

fig. 25. make OP =00. At P put in two op- 

posite and equal forces, S and T, equal and parallel to P'. 
The supposed equilibrium is undisturbed. But if P', P, and 



28 MECHANICS OF ENGINEERING. 

P are in equilibrium, so ought (by symmetry about 0) S, jP, 
and P to be in equilibrium, and they may be removed without 
disturbing equilibrium. But we have left Tand P', which are 
evidently not in equilibrium; .*. the proposition is proved by 
this reductio ad absurdum. Conversely a couple has no single 
resultant. 

30. A couple may he transferred anywhere in its own plane. 
— First, it may be turned through any angle a, about any 

PI point of its arm, or of its arm produced. 

G F™"7 ]-£ -i" Let (P, P')be a couple, G any point of its 

V''' 'fi'i t p ' arm (produced), and a any angle. Make 

^JU^dx GG — GA, CD — AB, and put in at G, 

? \ \ 2 \ | P 1 and P 2 equal to P (or P'), opposite to 

D\ Pt — --k;''*[ each other and perpendicular to GG', and 

3 \ ff P z and P 4 similarly at D. Now apply and 

fig. 26. combine P and P 1 at G, P' and P 4 at G'; 

then evidently It and R' neutralize each other, leaving P z and 
P 2 equivalent to the original couple (P, P f ). The arm 
GD = AB. Secondly, if G be at infinity, and a = 0, the 
same proof applies, i.e., a couple may be moved parallel to 
itself in its own plane. Therefore, by a combination of the 
two transferrals, the proposition is established for any trans- 
ferral in the plane. 

31. A couple may he replaced hy another of equal moment 
in a parallel plane. — Let (P, P') be a couple. Let GD, in a 
parallel plane, be parallel to AB. At D put in a pair of equal 

AE 

and opposite forces, S z and S„ parallel to P and each = =JP. 

ED 



RW 

Similarly at G, S, and £, parallel to P and each = ==^=^« 

IlC 

But, from similar triangles, 

^ l ±M. • <?_<? e c 
prj) — ]?£}•> ' ' °\ — °i — -° s — °4- 



STATICS OF A RIGID BODY. 29 

[Note. — The above values are so chosen that the intersection point E 
may be the point of application of (P' -\- S a ), the resultant of P and & a ; 
and also of (P-J-^s), the resultant of Pand &, as follows from § 21; thus 
(Fig. 28), P, the resultant of the two parallel forces Pand ti 3 , is = P-\-S 3 , 
and its moment, about any centre of moments, as E, its own point of ap- 
plication, should equal the (algebraic) sum of the moments of its com- 

AM 

ponents about E; i.e., R X zero = P. AE — S 3 . DE\ . \ S 3 = ==• P.] 



BE 



< P+S aH A 

1a 



P 

cr 



Lr"W t hi R4 

1 p fB. pi 

Fig. 27. Fig. 28. 

Replacing P' and £ 2 by {!>' + £ 2 ), and P and £ 3 by 
(jP -f~ S t ), the latter resultants cancel each other at E, leaving 
the couple (S v S 4 ) with an arm CD, equivalent to the original 

couple P, P' with an arm AB. But, since S 1 — === . P — 

jLB 

== . P, we have S 1 X CD = PxAB ; that is, their moments 

L>jD 

are equal. 

32. Transferral and Transformation of Couples. — In view of 

the foregoing, we may state, in general, that a couple acting on 
a rigid body may be transferred to any position in any parallel 
plane, and may have the values of its forces and arm changed 
in any way so long as its moment is kept unchanged, and still 
have the same effect on the rigid body (as to rest or motion,, 
not in distorting ir). 

Corollaries. — A couple may be replaced by another in any 
position so long as their axes are equal and parallel and simi- 
larly situated with respect to their planes. 

A couple can be balanced only by another couple whose axis 
is equal and parallel to that of the first, and dissimilarly situ- 
ated. For example, Fig. 29, Pa being = Qb, the rigid body 
AB (here supposed without weight) is in equilibrium in each 



30 



MECHANICS OF ENGINEERING. 



<jase shown. By "reduction of a couple to a certain arm a" 
is meant that for the original couple whose arm is a', with 
forces each = P' , a new couple is substituted whose arm shall 
be = a, and the value of whose forces P and P must be com- 
puted from the condition 

Pa = P f a\ i.e., P = Pa' 



a. 




Fig. 29. 



Fig. 30. 



33. Composition of Couples.— Let (P, P r ) and (Q, Q') be two 

couples in different planes reduced to the same arm AB = a, 
which is a portion of the line of intersection of their planes. 
That is, whatever the original values of the individual forces 
and arms of the two couples were, they have been transferred 
and replaced in accordance with § 32, so that P.AB. the 
moment of the first couple, and the direction of its axis, p, 
have remained unchanged ; similarly for the other couple. 
Combining P with Q and P' with Q\ we have a resultant 
couple (7?, P f ) whose arm is also AB. The axes p and q of 
the component couples are proportional to P . AB and Q . AB, 
i.e., to P and Q, and contain the same angle as P and Q. 
Therefore the parallelogram p . . . q is similar to the parallelo- 
gram P . . . Q ; whence p : q : r\\P : Q : P, or p : q : r : : 
Pa : Qa : Pa. Also r is evidently perpendicular to the plane 
of the resultant couple (i?, P f ), whose moment is Pa. Hence 
r, the diagonal of the parallelogram on p and q, is the axis of 
the resultant couple. To combine two couples, therefore, we 
have only to combine their axes, as if they were forces, by a 
parallelogram, the diagonal being the axis of the resultant 
couple ; the plane of this couple will be perpendicular to the 



STATICS OF A RIGID BODY. 31 

axis just found, and its moment bears the same relation to the 
moments of the component couples as the diagonal axis to the 
two component axes. Thus, if two couples, of moments Pa 
and Qb, lie in planes perpendicular to each other, their result- 
ant couple has a moment Re = \Z(Paf -\- (Qb) 2 - 

If three couples in different planes are to be combined, the 
axis of their resultant couple is the diagonal of the parallelo- 
piped formed on the axes, laid off to the same scale and point- 
ing in the proper directions, the proper direction of an axis 
being away from the plane of its couple, on the side from 
which the couple appears of positive rotation. 

34. If several couples lie in the same plane their axes are 
parallel and the axis of the resultant couple is their algebraic 
sum ; and a similar relation holds for the moments : thus, in 
Fig. 24, the resultant of the two couples has a moment = Qb 
— Pa, which shows us that a convenient way of combining 
couples, when all in one plane, is to call the moments positive 
or negative, according as the ideal rotations are against, or with, 
the hands of a watch, as seen from the same side of the plane ; 
the sign of the algebraic siam will then show the ideal rotation 
of the resultant couple. 

35. Composition of Non-concurrent Forces in a Plane. — Let 

P v P % , etc., be the forces of the system ; x v y v x» y„ etc., the 



Xf 



/A \ 
-X\ -J^-x, J 

X /% x 24 --^r 



\C 



+ Y,' ~^x 

Fig. 31. Fig. 32. 



co-ordinates of their points of application ; and or,, a„ . . . etc., 
their angles with the axis X. Heplace P x by its components 
X x and Y x , parallel to the arbitrary axes of reference. At the 
origin put in two forces, opposite to each other and equal and 
parallel to X x ; similarly for Y r (Of course X x == P x cos a and 
Y~ x = P x sin a.) We now have P x replaced by two forces X l 



32 MECHANICS OF ENGINEERING. 

and Y 1 at the origin, and two couples, in the same plane, whose- 
moments are respectively — X 1 y l and + Yjo x , and are there- 
fore (§34) equivalent to a single couple, in the same plane with, 
a moment = (Y^—X^.). 

Treating all the remaining forces in the same way, the whole- 
system of forces is replaced by 

the force 2(X) —X x + X a + . . . at the origin, along the axis X\ 
the force 2( Y) = Y t + Z, + . . . at the origin, along the axis Y\ 

and the couple whose mom. G = 2 ( Yx — Xy), which may be 

called the couple C (see Fig. 32), and may be placed anywhere 

in the plane. Now 2(X) and 2( Y) may be combined into a 

force B ; i.e., 

2X 

B = V (2X) 2 -+- ^ Yy and its direction-cosine is cos a — -^-. 

Since, then, the whole system reduces to C and i?, we must 

have for equilibrium B = 0, and G — ; i.e., for equilibrium. 

2X= 0,2Y=0, and 2(Yx-Xy) = 0. . eq. (1) 

If B alone = 0, the system reduces to a couple whose mo- 
ment is G = 2( Yx—Xy) ; and if G alone = the system re- 
duces to a single force B, applied at the origin. If, in general, 
neither B nor G = 0, the system is still equivalent to a single- 
force, but not applied at the origin (as could hardly be ex- 
pected, since the origin is arbitrary) ; as follows (see Fig. 33) : 

Replace the couple C by one of equal moment, G, with each 

C 
force = B. Its arm will therefore be -77. Move this couple 

in the plane so that one of its forces B may cancel the B al- 
ready at the origin, thus leaving a single resultant B for the- 
whole system, applied in a line at a perpendicular distance, 

c = -75 , from the origin, and making an angle a whose cosine = 
-75-, with the axis X. 

36. More convenient form for the equations of equilibrium 
of non-concurrent forces in a plane. — In (I.), Fig. 34, being 



STATICS OF A KIGTD BODY. . 33 

any point and a its perpendicular distance from a force P; 
put in at two equal and opposite forces P and P' = and || 
to P, and we have P replaced by an equal single force P' at 
0, and a couple whose moment is -|- Pa. (II.) shows a simi- 
lar construction, dealing with the JTand ^components of P, 
so that in (II.) P is replaced by single forces X' and Y' at 



Fig. 33. 



(and they are equivalent to a resultant P\ at 0, as in (I.), and 
two couples whose moments are -\- Yx and — Xy. 

Hence, being the same point in both cases, the couple Pa 
is equivalent to the two last mentioned, and, their axes being 
parallel, we must have Pa = Yx — Xy. Equations (1), 
§ 35, for equilibrium, may now be written 

2X = 0, 2 Y = 0,. and 2 {Pa) = 0. . . (2) 

In problems involving the equilibrium of non-concurrent 
forces in a plane, we have three independent conditions, or 
equations, and can determine at most three unknown quantities. 
For practical solution, then, the rigid body having been made 
free (by conceiving the actions of all other bodies as repre- 
sented by forces), and being in equilibrium (which it must be 
if at rest), we apply equations (2) literally ; i.e., assuming an 
origin and two axes, equate the sum of the ^components of 
all the forces to zero; similarly for the Y components ; and 
then for the "moment-equation," having dropped a perpen- 
dicular from the origin upon each force, write the algebraic 
sum of the products {moments) obtained by multiplying each 
force by its perpendicular, or "lever-arm" equal to zero, call- 
ing each product -j- or — according as the ideal rotation ap- 
pears against, or with, the hands of a watch, as seen from the 
same side of the plane. (The converse convention would do as 
well.) 



34- MECHANICS OF ENGINEERING. 

Sometimes it is convenient to use three moment equations, 
taking a new origin each time, and then the 2X ' = and 2 Y 
= are superfluous, as they would not be independent equa- 
tions. 

37. Problems involving Non-concurrent Forces in a Plane. — 

Remarks. The weight of a rigid body is a vertical force 
through its centre of gravity, downwards. 

If the surface of contact of two bodies is smooth the action 
(pressure, or force) of one on the other is perpendicular to the 
surface at the point of contact. If a cord must be imagined 
cut, to make a body free, its tension must be inserted in the 
line of the cord, and in such a direction as to keep taut the 
small portion still fastened to the body. In case the pin of 
a hinge must be removed, to make the body free, its pressure 
against the ring being unknown in direction and amount, it is 
most convenient to represent it by its unknown components X 
and Y, in known directions. In the following problems there 
is supposed to be no friction. If the line of action of an un- 
known force is known, but not its direction (forward or back 
ward), assume a direction for it and adhere to it in all the three 
equations, and if the assumption is correct the value of the 
force, after elimination, will be positive ; if incorrect, negative. 
Problem 1. — Fig. 35. Given an oblique rigid rod, with two 
loads G 1 (its own weight) and G t ; required the reaction of the 
smooth vertical wall at A, and the direction and amount of the 
Az'm^-pressure at 0. The reaction at A 
must be horizontal ; call it X' '. The pres- 
^-i sure at 0, being unknown in direction, will 
have both its X and Y components un- 
known. The three unknowns, then, are 
X , X\ and Y , while G^ G„ a,, a» and 
h are known. The figure shows the rod 
as a free body, all the forces acting on it 
have been put in, and, since the rod is at rest, constitute a sys- 
tem of non-concurrent forces in a plane, ready for the condi- 
tions of equilibrium. Taking origin and axes as in the figure. 



6 



Fig. 35. 



STATICS OF A RIGID BODY. 



35 



2X = gives +X - X' = ; 2Y = gives + T - G k 
— G 7 = ; while 2 (Pa) = 0, about 0, gives + X'h — 
G^a x — G 3 a t = 0. (The moments of X and Y about O 
are, each, = zero.) By elimination we obtain Y = G x -f- 
\G x a x A- G 2 a^\ ~ h; while the pressure at 
\ and makes with the horizontal an angle 

r-X. 



x = x 



# 2 ; 

o = vx; + r 

whose tan = Y n - 



[N.B. A special solution for this problem consists in this, that the result- 
ant of the two known forces 6r x and G 2 intersects the line of X! in a point 
which is easily found by § 21. The hinge-pressure must pass through this 
point, since three forces in equilibrium must be concurrent.] 

We might vary this problem by limiting X to a safe value, 
depending on the stability of the wall, and making h an un- 
known. The three unknowns would then be X , Y , and h. 

Problem 2. — Given two rods with loads, three hinges (or 
u pin -joints"), and all dimensions: required the three hinge- 

f y i 





Fig. 36. 



Fig. 37. 



pressures; i.e., there are six unknowns, viz., three JTand three 
Y components. We obtain three equations from each of the 
two free bodies in Fig. 37. The student may fill out the de- 
tails. Notice the application of the principle of action and 
reaction at B (see § 3). 

Problem 3. — A Warren bridge-truss rests on the horizontal 
smooth abutment-surfaces in Fig. 38. It is composed of equal 
isosceles triangles ; no piece is 
continuous beyond a joint, each 
of which is a pin connection. All 
loads are considered as acting at 
the joints, so that each piece will 
be subjected to a simple tension 
or compression. 




Fig. 38. 



36 



MECHANICS OF ENGINEERING. 



First, required the reactions of the supports Y x and F~ 2 ; 
these and the loads are called the external forces. 2(Pa) 
about = gives (the whole truss is the free body) 

Vfia — P x .\a — P 2 .§a — P z .\a = ; 

while 2(Pa) about K = gives 



and 



V x . 3a + P s . ia + Pjjp + P^a = ; 
V, = *[5A + 3P 2 + P 3 ] ; 
V, = il^ + 3P 2 + 5PJ. 



Secondly, required the stress (thrust or pull, compression or 
tension) in each of the pieces A, P, and Cent by the imaginary 
line PP. The stresses in the pieces are called "W^/wV ^-ces. 
These appear in a system of forces acting on a free body only 
when a portion of the truss or frame is conceived separated 
from the remainder in such a way as to expose an internal 
plane of one or more pieces. Consider as a free body the por- 
tion on the left of DE (that on the right would serve as well,. 
| p p but the pulls or thrusts in A, P, and 

C would be found to act in directions 
opposite to those they have on the 
other portion ; see § 3). Fig. 39. The 
p_£J>io arrows (forces) A* P, and C are not 

■ a ■* pointed yet. They, with V x , P x , and 

FlG - 39 - P 2 , form a system in equilibrium. 

2(Pa) about = gives 




-i* H 



(Ah) - Vfia + P x . %a + P 2 



0. 



Therefore the moment (Ah) = £a[4 V x — SP X — PJ, which 
is positive, since (from above) 4 Y x is > 3P X -\- P v Hence 
A must point to the left, i.e., is a thrust or compression, and is 

Similarly, taking moments about O x , the intersection of A 
and P, we have an equation in which the only unknown is (7, 
V&a + P x a = 0. .-. (Ch) = ia[3 V x - 2P,], 



viz., (Ch) 



STATICS OF A EIGID BOBT- 



37 



a positive moment, since 3 V 1 is >2P, ; .-. C must point te the 

right, i.e., is a tension, and = <w-[3 F, — 2PJ. 

Finally, to obtain B, put -^(vert. comps.) = 0; i.e. (i> cos <p) 
+ V x -P } - P 2 = 0. .:B cos <p = P l + P 2 -F l5 but 

(see foregoing value of T^) we may write 

F, = (P, + P 2 ) - (^P, + iP 2 ) + iP s . 
„\ i? cos 9? will be + (upward) or — (downward), and B will 
be compression or tension, as \P Z is < or > [\P X -f- JPJ. 

P = [P,+ P 2 - FJ + cos ^ = JL ^- 4 - [P, + i>. - FJ. 

Problem 4. — Given the weight £r a of rod, the weight G 23 

and all the geometrical elements (the student will assume a 

W\|P, 





Fig. 40 



Fig. 41. 

convenient notation); required the tension in the cord, and the 
amount and direction of pressure on hinge-pin. 

Problem 5. — Roof-truss ; pin-connection ; all loads at joints ; 
wind-pressures W and TF, normal to OA ; required the three 
reactions or supporting forces (of the two horizontal surfaces 
and one vertical surface), and the 



stress in each piece. All geomet- 



rical elements are given ; 
P„ P 2 , TT(Fig.>0). 



also P, 



38. Composition of Non-concur- 
rent Forces in Space. — Let P„ P„ 

etc., be the given forces, and x x , y^ 
2j, x„ y„ z» etc., their points of ap- 
plication referred to an arbitrary 
origin and axes ; or,, /3„ y x , etc., 
the angles made by their lines of application with X_ J^and Z. 




38 



MECHANICS OF ENGINEERING. 



Considering the first force P x , replace it by its three com- 
pon en ts parallel to the axes, X x = P x cos a x \ Y x = P 1 cos fi x \ 
and Z x = P 1 cosy l (P x itself is not shown in the figure). At 
0, and also at A, put a pair of equal and opposite forces, 
each equal and parallel to Z x \ Z x is now replaced by a single 
force Z x acting upward at the origin, and two couples, one 
in a plane parallel to YZ and having a moment = — Z x y x (as 
we see it looking toward from a remote point on the axis 
-\- X), the other in a plane parallel to XZ and having a mo- 
ment = -f- Zjc x (seen from a remote point on the axis -f~ Y). 
Similarly at and (7 put in pairs of forces equal and parallel 
to X x , and we have X„ at _B, replaced by the single force X x 
at the origin, and the couples, one in a plane parallel to XY X 
and having a moment -f- X x y x , seen from a remote point on 
the axis ~\- Z, the other in a plane parallel to XZ, and of a 
moment = — X X 2 X , seen from a remote point on the axis -f- Y\ 
and finally, by a similar device, Y x at B is replaced by a force 
Y x at the origin and two couples, parallel to the planes XY 
and YZ, and having moments — Y t x x and -|- Y x z^ respective- 
ly. (In Fig. 42 the single forces at the origin are broken 
lines, while the two forces constituting any one of the six 

couples may be recognized as being 
equal and parallel, of opposite di- 
rections, and both continuous, or 
both dotted.) We have, therefore, 
replaced the force P x by three 
forces X^ Y„ Z y , at 0, and six 
couples (shown more clearly in 
Fig. 43; the couples have been 
transferred to symmetrical posi- 
tions). Combining each two couples 
whose axes are parallel to X, Y, 
or Z, they can be reduced to three, viz., 

one with an X axis and a moment = Y x z x — Z x y x ; 
one witli a ]Taxis and a moment = Z x x x — X 1 s l ; 
one with a Z axis and a moment = X,y 1 — Y x x x . 




Fig. 43. 



STATICS OF A RIGID BODY. 39 

Dealing with each of the other forces jP 2 , P z , etc., in the same 
manner, the whole system may finally be replaced by three 
forces 2X, 2 Y, and 2Z, at the origin and three couples 
whose moments are, respectively, 

L = 2( ~Yz — Zy) with its axis parallel to X; 
M = 2(Zx — Xz) with its axis parallel to Y; 
X = ^{Xy — Yx) with its axis parallel to Z. 

The "axes" of these couples, being parallel to the respective 
co-ordinate axes X, Y, and Z, and proportional to the mo- 
ments I, M, and X, respectively, the axis of their resultant 
C, whose moment is G, must be the diagonal of a parallelo- 
pipedon constructed on the three component axes (propor- 
tional to) Z, M, and X. Therefore, G = VL 2 -\-M 2 + X% 
while the resultant of J£X. 2 Y. and 2Z is 



r = V(zxy + (2 yj + {zzy 

acting at the origin. If <*, /3, and y are the direction-angles 

of R, we have cos a — -^- cos p = ~jj-, and cos y = -=: ; 

it, ii, J£ 

while if A, ju, and v are those of -the axis of the couple C, we 

L M , X 

have cos X = jy, cos jx = -o, and cos r = y 7 . 

For equilibrium we have both G = and i? = ; i.e., 
separately, six conditions, viz., 

2X-= 0, 2 Y = 0, 2Z=0 ; and Z=0, Jf=0, ir=0 . (1) 

Now, noting that ^X = 0, 2Y = 0, and ^(Xy - Pb) = 
are the conditions for equilibrium of the system of non-concur- 
rent forces which would be formed by projecting each force of 
our actual system upon the plane XY, and similar relations 
for the planes YZ and XZ, we may restate equations (1) in 
another form, more serviceable in practical problems, viz. : 
Note. — If a system of non-concurrent forces in space is in 
equilibrium, the plane systems formed by projecting the given 
system upon each of three arbitrary co-ordinate planes will each 
be in equilibrium. But we can obtain only six independent 



40 MECHANICS OF ENGINE K RING. 

equations in any case, available for six unknowns. If B alone 
= 0, we have the system equivalent to a couple G, whose 
moment = G ; if G alone == 0, the system has a single re- 
sultant B applied at the origin. In general, neither B nor G 
being = 0, we cannot further combine B and (as was done 
with non-concurrent forces in a plane) to produce a single re- 
sultant unless B and G are in the same plane ; i.e., when the 
angle between B and the axis of 6" is = 90°. Call that angle 
6. If, then, cos 6 = cos a cos A -|- cos ft cos yw -f- cos y cos v 
is = = cos 90°, we may combine B and C to produce 2 
single resultant for the whole system ; acting in a plane con- 
taining B and parallel to the plane of G in a direction parallel 

C 
to B, at a perpendicular distance g = -^ from the origin and 

= B in intensity. The condition that a system of forces in 
space have a single resultant is, therefore, substituting the 
previously derived values of the cosines, (2JE) . L-\-{2Y) . M 
+ (2Z) .JV=Q. 

This includes the cases when B is zero and when the system 
reduces to a couple. 

To return to the general case, B and G not being in the 

same plane, the composition of forces in space cannot be 

further simplified. Still we can give any value we please to 

jP, one of the forces of the couple G, calculate the correspond- 

G 
ino; arm a = -7^, then transfer G until one of the P's has the 

same point of application as B, and combine them by Hie 
parallelogram of forces. We thus have the whole system 
equivalent to two forces, viz., the second B, and the resultant 
of B and the first P. These two forces are not in the same 
plane, and therefore cannot be replaced by a single resultant. 

39. Problem. (Non-concurrent forces in space.) — Given all 
geometrical elements (including a, /?, y, angles of P\ also the 
weight of Q, and weight of apparatus G ; A being a hinge whose 
pin is in the axis Y, Ga ball-and-socket joint: required the 
amount of P (lbs.) to preserve equilibrium, also the pressures 



STATICS OF A RIGID BODY. 



41 



{amount ana direction) at A and ; no friction. Beplace P 
by its X, Y, and Z components. The pressure at A will have 




Fig. 44. 



Z and X components ; that at 0, X, Y, and Z components. 
The body is now free, and there are six unknowns. 
2X, 2 Y, and 2Z give, respectively, 

P cos a + X x + X = ; 

P cos fi -f y = ; and Z, + Z — Q — G - P cos y = 0. 

As for moment-equations (see note in last paragraph), project- 
ing the system upon YZ and putting 2 (Pa) about = 0, 
we have 

-Z^+^+ &? + (Pcos r )& + (Pcos/fy? = 0; 
projecting it upon XZ, and putting 2 (Pa) about = 0, we 
have $7* — (i 3 cos a)c — (P cos ^)a = ; 

projecting on XY, moments about give 

X x l + (P cos a)5 - (P cos /?)a = 0. 

From these six equations we may obtain the six unknowns, 
P, X , Y , Z , X^ and Z x . If for any one of these a negative 
result is obtained, it shows that its direction in Fig. 4A should 
be reversed. 



42 MECHANICS OF ENGINEERING. 



CHAPTER IV. 

STATICS OF FLEXIBLE CORDS. 

40. Postulate and Principles. — The cords are perfectly flexi- 
ble and inextensible. All problems will be restricted to one 
plane. Solutions of problems are based on three principles, 
viz.: 

Prin. I. — The strain on a cord at any point can act only 
along the cord, or along the tangent if it be curved. 

Prin. II. — We may apply to flexible cords in equilibrium all 
the conditions for the equilibrium of rigid bodies ; since, if the 
system of cords became rigid, it would still, with greater rea- 
son, be in equilibrium. 

Pkin. III. — The conditions of equilibrium cannot be applied, 
of course, unless the system can be considered a free body, 
which is allowable only when we conceive to be put in, at the 
points of support or fastening, the reactions (upon the cord) 
of those points and the supports removed. These reactions 
having been put in, then consider the case in Fig. 45 in one 
plane. If we take any point, p, on the cord as a centre 
of moments, knowing that the resultant P, of the forces P v 
P 3 , and P % , situated on one side oip, must act along the cord 

iP 6 through p (by Prin. 1), therefore 
we have P 1 a 1 — P 2 # 2 — P 3 # 3 
= R X zero — 0, and (equally 
well) P G a 6 — P b a b — P^a, = 0. 
That is, in a system of cords in 
FlG> 45> equilibrium in a plane, if a centre 

of moments be taken on the cord, the algebraic sum of the mo- 




STATICS OF FLEXIBLE CORDS. 



43 



ments of those forces situated on one side {either) of this point 
will equal zero. 

41. The Pulley. — A cord in equilibrium over a pulley whose 
axle is smooth has the same tension on both sides ; for, Fig. 46, 

C; 





Fig. 46. 



Fig. 47. 



considering the pulley and its portion of cord free 2(Pa) = 
about the centre of axle gives P'r — Pr, i.e., P' — P = ten- 
sion in the cord. Hence the pressure P at the axle bisects 
the angle a, and therefore if a weighted pulley rides upon a 
cord ABC, Fig. 47, its position of equilibrium, B, may be 
found by cutting the vertical through A by an arc of radius 
CD = length of cord, and centre at C, and drawing a horizon- 
tal through the middle of AD to cut CD in B. A smooth 
ring would serve as well as the pulley ; this would be a slip- 
knot. 



42. If three cords meet at & fixed ~knot, and are in equilib- 
rium, the tension in any one is the equal and 
opposite of the resultant of those in the other 
two. 

43. Tackle. — If a cord is continuous over a 
number of sheaves in blocks forming a tackle, 
neglecting the weight of the cord and blocks and 
friction of any sort, we may easily find the ratio 
between the cord-tension P and the weight to be 
sustained. E.g., Fig. 48, regarding all the straight 
cords as vertical and considering the block B 
free, we have. Fig. 49 (from 2Y= 0), 4P — G 

G ^ 




= 0,.:P 



4' 



The stress on the support C will = 5P. 



44 



MECHANICS OF ENGINEERING. 




G 2 
Fig. 50, 

similarly H n and 



44. Weights Suspended by Fixed Knots. — Given all the geo- 
metrical elements in Fig. 50, and 
one weight, G x \ required the re- 
maining weights and the forces 
IT , V , H n and V n , at the points 
of support, that equilibrium may 
obtain. JI and V are the hori- 
zontal and vertical components of 
the tension in the cord at 0; 
V n those at n. There are n -f- 2 unknowns. 
From Prin. II we have 2X = 0, and 2 Y = 0; i.e., B - H n 
= 0, and [G x 4- G 2 + . . . ] - [ V + V n ] '= 0. While from 
Prin. III., taking the successive knots, 1, 2, etc., as centres of 
moments, we have 

— Vox, + R €l = 0, 

— Vox, + H,y 2 + Gfa — x x ) = 0, 

— V Q x, + H Q y, + G x {x z - x,) + G,(x 3 - x 2 ) = 0, 

etc., for n knots. 

Thus we have n -f- 2 independent equations, a sufficient 
number, and they are all of the first degree (with reference to 
the unknowns), and easily solved. As a special solution, we 
may, by § 42, resolve G, in the directions of the first and sec- 
ond cord-segments, and obtain their tensions by a parallelogram 
of forces ; then at the second knot, knowing the tension in the 
second segment, we may find that in the third and G 9 in like 
manner, and so on. Of course JT and V are components of 
the tension in the first segment, H n and V n of that in the 
last. 



.45. The converse of the problem in §44, viz., given the 
weights 6r x , etc., x n and y n , the lengths a, b, c, etc.; required 
iZo, Fo, H n . V w and the co-ordinates x v y„ a? 2 , y a . etc., of the 
fixed knots when equilibrium exists, contains 2n -f- 2 un- 
knowns. Statics furnishes n -\- 2 equations (already given in 
§44); while geometry gives the other n equations, one for 
•each cord-segment, viz., x* -f* Vi — a * j (^a ~~ x iY 4" (SVr - V^ 
= If ; etc. 



STATICS OF FLEXIBLE CORDS. 



45 



However, most of these 2n-{-2 equations are of the second 
degree ; hence in the general case they cannot be solved. 

46. Loaded Cord as Parabola. — If the weights are equal and 
infinitely small, and are intended to be uniformly spaced 
along the horizontal, when equilib- 
rium obtains, the cord having no 
weight, it will form a parabola. Let 
a = weight of loads per horizontal 
linear unit, O be the vertex of the 
curve in which the cord hangs, and 
m any point. We may consider 
the portion 0?n as a free body, if 
the reactions of the contiguous portions of the cord are put in,. 
Hq and T, and these (from Prin. I.) must act along the tangents 
to the curve at and m, respectively ; i.e., H Q is horizontal, 

and T makes some angle cp (whose tangent = y-, etc.) with 
the axis X. Applying Prin. IP, 

(1HR 

2X = gives Tcoscp-B =0; i.e., T T = B ; 




ds 



qx. 



(i) 

(2) 



2 Y*= gives T sin cp — qx 

Dividing (2) by (1), member by member, we have ~ = ~j ; 

q 
:. dy = -jj-xdx, the differential equation of the curve ; 

y = ~tj I x ^ x — ~tt • o" ; or » 2 = y, the equation of a 

parabola whose vertex is at and axis vertical. 

Note. — The same result, -^- = ~, mav be obtained by considering that 

dx IJq - J & 

we have here (Prin. II.) a free rigid body acted on 

by three forces, T, Bo, and R = qx, acting verti- 

'" cally through the middle of the abscissa x\ the 

resultant of Bq and R must be equal and oppo- 




site to T, Fig. 52. 



R dy qx 
tan cp = — , or ~ = \ T . 
^ B dx B 



Evidently also the tangent-line bisects the ab- 
scissa x. 



46 MECHANICS OF ENGINEERING. 

47. Problem under § 46. [Case of a suspension-bridge in 
which the suspension-rods are vertical, the weight of roadway 
is uniform per horizontal foot, and large compared with that 
of the cable and rods. Here the roadway is the only load : it 
is generally furnished with a stiffening truss to avoid deforma- 
tion under passing loads.] — Given the span == 2&. Fig. 53, 
Y| v f 71 the deflection = a, and the rate of loading 

j. j) — -■^fii* = q lbs. per horizontal foot ; required the 

a — x--prf I 1 tension in the cable at 0, also at m : and 

^pj . r'iT^T't the length of cable needed. From the 

fig. 53. equation of the parabola qx 2 = 2H$j, put- 

ting x = b and y = a, we have H = qlf -r- 2a = the tension 
at 0. From 27=0we have V 1 = #&, while ^X= gives 

1 



H x = E ; .-. the tension at m = \/ H? + V?= ^~{qb V4a a + 6 2 ]. 

Za 

The semi-length, (9m , of cable (from p. 88, Todhunter's In- 
tegral Calculus) is (letting n denote H ~ 2q, = I? -f- 4#) 

Om — Vna -\- a 2 -\- n . log e [( Va-\- Vn -\- a) ~- V^]. 

48. The Catenary.* — A flexible, in extensible cord or chain, of 
uniform weight per unit of length, hung at two points, and 
supporting its own weight alone, forms a curve called the 
eatenary. Let the tension H Q at the lowest point or vertex be 
represented (for algebraic convenience) by the weight of an 
imaginary length, c, of similar cord weighing q lbs. per unit 
of length, i.e., H = qc\ an actual portion of the cord, of 
length s, weighs qs lbs. Fig. 54 shows us free and in equilib- 
rium a portion of the curve of any 
length s, reckoning from the 
vertex. Required the equation of 
the curve. The load is uniformly 
spaced along the curve, and not 
horizontally, as in §§ 46 and 47. 

fig. 54. 2 Y = gives 1~ = qs ; while 

2X = gives T-j- = qc. Hence, by division, cdy = sdx, and 
squaring, &dy l — s^dx* (1) 

* For the " transformed catenary," see p. 395. 




STATICS OF FLEXIBLE CORDS. 4? 

Put dy 2 = ds 2 — dx 2 , and we have, after solving for dx 
cds /»s ds p 

and a* =o . log.[(5 + V7+?) -J- <?], ... (2) 

a relation between the horizontal abscissa and length of curve. 
Again, in eq. (1) put dx 2 = ds 2 — dy 2 , and solve for dy. 

This gives ^ = -=_ = j . -^p^- Therefore 
y = *e/V + *T*^ + = *["! 2 ^ + ^ and finally 



y = Ys 2 + & — e (3) 

Clearing of radicals and solving for c, we have 

c = (s*-f) + 2y (4) 

Example. — A 40-foot chain weighs 240 lbs., and is so hung 
from two points at the same level that the deflection is 10 
feet. Here, for s = 20 ft., y = 10 ; hence eq. (4) gives the 
parameter, c = (400 — 100) -*■ 20 = 15 feet. ^ = 240 -^ 40 
= 6 lbs. per foot. .*. the tension at the middle is IT = qc 
= 6 X 15 = 90 lbs.; while the greatest tension is at either 
support and = VdO 2 + 120 s = 150 lbs. 

Knowing e = 15 feet, and putting s = 20 feet = half 
length of chain, we may compute the corresponding value of 
x from eq. (2) ; this will be the half-span [log e m = 2.30258 
X (common log m)]. To derive s in terms of x, transform 
eq. (2) in the sense in which n = log e m may be transformed 
into e n = m, clear of radicals, and solve for s, which gives 



= *>[< - 



(5) 



Again, eliminate s from (2) by substitution from (3), trans- 
form as. above, clear of radicals, and solve for y -j- c, whence 

y+e = tc\f + f*], (6) 



48 MECHANICS OF ENGINEERING. 

which is the equation of a catenary with axes as in Fig. 54. 
If the horizontal axis be taken a distance = c below the ver- 
tex, the new ordinate y' = y -f- c, while x remains the same ; 
the last equation is simplified. 

If the span and length of chain are given, or if the span 
and deflection are given, c can be determined from (5) or (6) 
only by successive assumptions and approximations. 

48a. Addendum to § 55. Mass. — In Physics, the fundamental units are 
those of 

Space, involving a unit of length (and thence of area and volume) ; 
Time, " a unit of time, usually the second ; 

Mass, " a unit of mass, which (by Government decree) maybe the 

quantity of matter in a specified piece of platinum, or specified volume of water, 
etc. (a beam-balance being used to determine equal quantities of mass) ; while 
Force involves a derived unit, being measured by its effect in accelerating 
the velocity of a moving mass, since it is proportional both to the mass and the 
acceleration. The unit force (called absolute unit) is the force necessary to pro- 
duce unit acceleration in a unit of mass; so that to produce an acceleration = p in 
a mass = m requires a force = F— inp, and the force thus obtained is in absolute 
units. This is called the dynamic measure of a force. 

Example. — In the C.G.S. system of units, required the constant force necessary 
to cause a mass of 400 grams to gain 200 velocity units in 2 sec ; i.e., p = 100' 
centims. per sec, per sec. From P' = nip we have 

F- 400 x 100 = 40000 abs. units of force (or dynes, in C.G.S. system). 
In the ft. -lb. -sec. system the absolute unit is called apoundal. 

In Mechanics of Engineering, however, it is more convenient to regard the- 
fundamental units to be those of 

Space, as ft., metre, etc., area and volume corresponding ; 

Time, as seconds, hours, etc. ; 

Force, as lbs., grams, kilograms, tons, etc., indicated by a spring balance; 
while for 

Mass we assume a derived unit, a mode of measuring it being developed as- 
follows : 

If by experiment (block on smooth table, for instance) we find that a constant 
force P (lbs., tons, kilos.) will maintain an acceleration -p in the rectilinear- 
motion (in line of force) of a body whose weight (by previous trial with a spring 
balance) is G (lbs. , tons, or other unit) ; and if in a second experiment, by allow- 
ing the force G to act on the same body in vacuo, a free vertical fall with accelera- 
tion = g is the result,— we find that the proportion (Newton's 2d Law) P: G ::p:g 

is verified. This may be written P= — . p, and may then be read: Force = mass 

X acceleration, if we call the quotient G + g the Mass of the body whose weight 
(by spring balance) is = G at a locality where the acceleration of gravity = g ; for 
this quotient will be the same at all localities on the earth's surface. 

Example (same as above). — If a body whose weight G = 400 grams (force) is to ; 
have its velocity increased, in 2 sec, from 300 centims. per sec. to 500 centims. 
per sec, at a uniform rate, we must provide a constant force 

n 400 40000 

P= ggj X 100 = -^- = 40.77 grams; or .040 kilos. 

This is called the gravitation measure of a force. Hence it is evident that to re- 
duce absolute units (called dynes and poundals) in the C.G.S. and ft.-lb.-sec 
systems, respectively) to ordinary practical units of force (lbs., tons, kilos., etc.,. 
of a spring balance), we divide by the value of g proper to the system of units em- 
ployed ; and vice versd. 



PAKT II -DYNAMICS, 



CHAPTEE I. 

RECTILINEAR MOTION OF A MATERIAL POINT. 

49. Uniform Motion implies that the moving point passes 
over equal distances in equal times ; variable motion, that un- 
equal distances are passed over in equal times. In uniform 
motion the distance passed over in a unit of time, as one sec- 
ond, is called the velocity (= v), which maj also be obtained 
by dividing the length of any portion (= s) of the path by 
the time (= t) taken to describe that portion, however small or 
great ; in variable motion, however, the velocity varies from 
point to point, its value at any point being expressed as the 
quotient of ds (an infinitely small distance containing the 
given point) by dt (the infinitely small portion of time in 
which ds is described). 

49«. By acceleration is meant the rate at which the velocity 
of a variable motion is changing at any point, and may be a 
uniform acceleration, in which case it equals the total change 
of velocity between any two points, however far apart, divided 
by the time of passage ; or a variable acceleration, having a 
different value at every point, this value then being obtained 
by dividing the velocity-increment, dv, or gain of velocity 
in passing from the given point to one infinitely near to it, by 
dt, the time occupied in acquiring the gain.* (Acceleration 
must be understood in an algebraic sense, a negative accelera- 
tion implying a decreasing velocity, or else that the velocity in 
a negative direction is increasing.) The foregoing applies to 
motion in a path or line of any form whatever, the distances 
mentioned being portions of the path, and therefore measured 
along the path. 

I * See addendum on p. 12. 



50 MECHANICS OF ENGINEERING. 

50. Rectilinear Motion, or motion in a straight line.— The 
general relations of the quantities involved may be thus stated 
(see Fig. 55) : Let v = velocity of the body at any instant ; 

-6 0, 8—»a 8 d8 m +S then dv = gain of velocity 

~~* f \ \ | in an instant of time dt. Let 

idt*dt\ i t — time elapsed since the 
body left a given fixed point, 
which will be taken as an origin, 0. Let s = distance (+ or 
— ) of the body, at any instant, from the origin 0; then ds = 
distance traversed in a time dt. Let^> = acceleration == rate 
at which v is increasing at any instant. All these may be 
variable ; and t is taken as the independent variable, i.e., time 
is conceived to elapse by equal small increments, each = dt ; 
hence two consecutive ds's will not in general be equal, their 
difference being called d 2 s. Evidently dH is = zero, i.e., dt is 
constant. . 

Since -j- = number of instants in one second, the velocity at 
any instant (i.e., the distance which would be described at that % 
rate in one second) is 7) = ds . ^- ; .'. v = — (I.) 



dV " dt' 

anrl I : 



1 /. (l*\ **\ 

Similarly, p = dv . ^, and I since dv = d\ c i^J =~^r 



dv d*s /TT . 

■ , -V = -dt = df W 

Eliminating dt, we have also vdv =pds (HI.) 

These are the fundamental differential formulas of rectilinear 
motion (for curvilinear motion we have these and some in ad- 
dition) as far as kinematics, i.e., as far as space and time, is 
concerned. The consideration of the mass of the material 
point and the forces acting upon it will give still another rela- 
tion (see § 55). 

51. Rectilinear Motion due to Gravity. — If a material point 
fall freely in vacuo, no initial direction other than vertical 
having been given to its motion, many experiments have 



RECTILINEAR MOTION OF A MATERIAL POINT. 51 

shown that this is a uniformly accelerated rectilinear motion 
in a vertical line having an acceleration (called the accelera- 
tion of gravity) equal to 32.2 feet per square second, or 9.81 
metres per square second ; i.e., the velocity increases at this 
constant rate in a downward direction, or decreases in an up- 
ward direction. 

[Note. — By " square second " it is meant to lay stress on the fact that an 
acceleration (being = d*s -r- dt 2 ) is in quality equal to one dimension of 
length divided by two dimensions of time. E.g., if instead of using the 
foot and second as units of space and time we use the foot and the minute, 
g will = 32.2 X 3600; whereas a velocity of say six feet per second would 
= 6 X 60 feet per minute. The value of g = 32.2 implies the units foot 
and second, and is sufficiently exact for practical purposes.] 

52. Free Fall in Vacuo. — Fig. 56. Let the body start at 

with an initial downward velocity = c. The accelera- _s 

tion is constant and = -f- g. Reckoning both time and I 

distance (-)- downwards) from 0, required the values of *° 

the variables s and v after any time t. From eq. (II.), " o 

§ 50, we have -f- g = dv -=- dt ; .'. dv = gdt, in which the v s 

two variables are separated. I I 

pv pt r-» t~t 1 i 

Hence J c dv = gJ Q dt; i.e., \jd = g\J; or v — c = «j 

gt — ; and finally, v — c + gt (1) fio. 56. 

(Notice the correspondence of the limits in the foregoing 
operation ; when t = 0, v = -f- c.) 

From eq. (I.), § 50, v = ds -f- dt ; .*. substituting from (1), 
ds = (c -\- gt)dt, in which the two variables s and t are sepa- 
rated. 

•'• X d8 = c Jo dt + &X m '> Le -' [/ = G \J+ ^[o 2 ' 
or s=ct + igf (2) 

Again, eq. (III.), § 50, vdv = gds, in which the variables v 
and s are already separated. 



/v ps r~v r~s 

vdv = gj ds ; or iv* = g s; i.e., i(v* - c*) 

\—c 1—0 



g*, 



s== ^i~ ( s > 



52 MECHANICS OF ENGINEERING. 

If the initial velocity = zero, i.e., if the body falls from rest, 
eq. (3) gives s=^ndv= V2gk* [From the frequent re- 

currence of these forms, especially in hydraulics, —is called the 

"height due to the velocity v," i.e., the vertical height through 
which the body must fall from rest to acquire the velocity v ; 
while, conversely, ^2gh j s called the velocity due to the height 
or head h.~\ 

By eliminating g between (1) and (3), we may derive another 
formula between three variables, s, v, and t, viz., 

s = i(c + v)t. ...... (4) 

53. Upward Throw. — If the initial velocity were in an up- 
ward direction in Fig. 56 we might call it — c, and introduce it 
with a negative sign in equations (1) to (4), just derived ; but 
for variety let us call the upward direction -)-, in which case 
an upward initial velocity would = -\- c, while the acceleration 
= — g, constant, as before. (The motion is supposed confined 
within such a small range that g does not sensibly vary.) Fig. 

57. From p = dv -=- dt we have dv = — gdt and 

m6] v 



-j \V nv at 

From v = ds -T- dt, ds = edt — gtdt, 

Xs r*t r*t 

ds = ej dt - gj tdt ; or s = ct - igf. (2> 

S vdv = jpds gives / vdv = — gj ds, whence 



.r 1 

; 
Fig. £ 



tf tf 

\(tf - c') = - #8, or finally, s = —%z-- • (3)a 

And by eliminating g from (l)a and (S)a, 

* = 2 (o + v)t (4)a 

The following is now easily verified from these equations : 
the body passes the origin again (s = 0) with a velocity = — e, 
after a lapse of time = 2c -h g. The body comes to rest (for 

* In Hydraulics h is used instead of s. 



RECTILINEAR MOTION OF A MATERIAL POINT. 53 

an instant) (put v = 0) after a time = o ■— g, and at a distance 
s = c" 2 ^- 2g (" height due to velocity c") from <?. For £ > 
c -^ g, v is negative, showing a downward motion ; for t > 
2<? -f- <7, s is negative, i.e., the body is below the starting-point 
while the rate of change of v is constant and = — g at all 
points. 

54. Newton's Laws. — As showing the relations existing in 
general between the motion of a material point and the actions 
(forces) of other bodies upon it, experience furnishes the fol- 
lowing three laws or statements as a basis for dynamics : 

(1) A material point under no forces, or under balanced 
forces, remains in a state of rest or of uniform motion in a 
right line. (This property is often called Inertia.) 

(2) If the forces acting on a material point are unbalanced, 
an acceleration of motion is produced, proportional to the re- 
sultant force and in its direction. 

(3) Every action (force) of one body on another is always 
accompanied by an equal, opposite, and simultaneous reaction. 
(This was interpreted in § 3.) 

As all bodies are made up of material points, the results ob- 
tained in Dynamics of a Material Point serve as a basis for the 
Dynamics of a Rigid Body, of Liquids, and of Gases. 

55. Mass.* — If a body is to continue moving in a right line, 
the resultant force P at all instants must be directed along that 
line (otherwise it would have a component deflecting the body 
from its straight course). 

In accordance with Newton's second law, denoting by^> the 
acceleration produced by the resultant force (G being the 
body's weight), we must have the proportion P : G : : p : g ; 
i.e., 

P = — .p , orP = Jfy.. . (IY.) 

Eq. IY. and (I.), (II.), (III.) of § 50 are the fundamental 
equations of Dynamics. Since the quotient G -v- g is invaria- 

* See Addendum on p. 48. 



54 MECHANICS OF ENGINEERING. 

ble, wherever the body be moved on the earth's surface (G and 
g changing in the same ratio), it will be used as the measure 
of the mass M or quantity of matter in the body. In this way 
it will frequently happen that the quantities G and g will ap- 
pear in problems where the weight of the body, i.e., the force 
of the earth's attraction upon it, and the acceleration of gravity 
have no direct connection with the circumstances. No name 
will be given to the unit of mass, it being always understood 
that the fraction G ~ g will be put for M before any numeri- 
cal substitution is made. From (IV.) we have, in words, 
j accelerating force = mass X acceleration; 

1 also, acceleration -— accelerating force ~ mass. 

56. Uniformly Accelerated Motion. — If the resultant force is 
constant as time elapses, the acceleration must be constant (from 
eq. (IV.), since of course If is constant) and = P ~ M. The 
motion therefore will be uniformly accelerated, and we have 
only to substitute + ]) (constant) for g in eqs. (1) to (4) of 
§ 52 for the equations of this motion, the initial velocity being 
= c (in the line of the force). 

v = c+pt ... (1); s=ct + ipf; . . . (2) 

* = ^r~ ); - ■ • (8) ' ««*« = *(»+«>)* • • ■ w 

If the force is in a negative direction, the acceleration will 
be negative, and may be called a retardation/ the initial veloc- 
ity should be made negative if its direction requires it. 

57. Examples of Unif. Ace. Motion. — Example 1. Fig. 58. 
A small block whose weight is -J lb. has already described a 
— S P M v B distance Ao = 48 inches over a 

! A SM00TH ~ = $ I1 ~^ — iSif — ~~| smooth portion of a horizontal 
fig. 58. table in two seconds ; at it en- 

counters a rough portion, and a consequent constant friction of 

2 oz. Required the distance described beyond 0, and the time 
occupied in coming to rest. Since we shall use 32.2 for g, 
times must be in seconds, and distances in feet ; as to the unit 



RECTILINEAR MOTION OF A MATERIAL POINT. 55 

of force, as that is still arbitrary, say ounces. Since AO was 
smooth, it must have been described with a uniform motion 
(the resistance of the air being neglected); hence with a veloc- 
ity == 4 ft. -r 2 sec. = 2 ft. per sec. The initial velocity for 
the retarded motion, then, is c = -f- 2 at 0. At any point be- 
yond the acceleration = force ■*- mass = (— 2 oz.) -f- (8 oz. 
-~ 32.2) = — 8.05 ft. per square second, i.e., p = — 8.05 = 
constant ; hence the motion is uniformly accelerated (retarded 
here), and we may use the formulae of § 56 with c = -f- 2, p = 
— 8.05. At the end of the motion v must be zero, and the 
corresponding values of s and t may be found by putting v = 
in equations (3) and (1), and solving for s and t respectively : 
thus from (3), s =£(-4)-r- (— 8.05), i.e., s = 0.248 +, which 
must be feet ; while from (1), t = (— 2) -r- (— 8.05) = 0.248 +, 
which must be seconds. 

Example 2. (Algebraic.) — Fig. 59. The two masses M x — 
G 1 -r- g and M = G ~ g are connected by a flexible, inexten- 
sible cord. Table smooth. Required the acceleration common 
to the two rectilinear motions, and the tension in- the string S, 






iu G N t 

Fig. 59. Fig. 60. 

there being no friction under G^ none at the pulley, and no 
mass in the latter or in the cord. At any instant of the mo- 
tion consider G 1 free (Fig. 60), N being the pressure of the 
table against G,. Since the motion is in a horizontal right line 
^(vert. compons.)= 0, i.e., JV— G x = 0, which determines iV". 
S, the only horizontal force (and resultant of all the forces) = 
M,p, i.e., 

S = G x p h- g (1) 

At the same instant of the motion consider G free (Fig. 61); 
the tension in the cord is the same value as above = S. The 
accelerating force is G — S, and 

.-. = mass X ace, or G — S = (G -f- g)p. . (2) 



V 



56 MECHANICS OF ENGINEERING. 

I a s From equations (1) and (2) we obtain p = (Gg) -~ 

l A (G -f- G x ) = a constant ; hence each motion is uniformly 

r S accelerated, and we may employ equations (1) to (4) of 

§ 56 to find the velocity and distance from the starting- 

* points, at the end of any assigned time t, or vice versa, 

' The initial velocity must be known, and may be zero. 

Also, from (1) and (2) of this article, 

8 = (GG,) + (G+ G x ) = constant 

Example 3. — A body of 2f (short) tons weight is acted on 
during \ minute by a constant force P. It had previously de- 
scribed 316f yards in 180 seconds under no force ; and subse- 
quently, under no force, describes 9900 inches in -fa of an hour. 
Kequired the value of P. Ans. P = 22.1 lbs. 

Example 4. — A mass of 1 ton weight, having an initial 
velocity of 48 inches per second, is acted on for J minute by a 
force of 400 avoirdupois ounces. Required the final velocity. 

Ans. 10.037 ft. per sec. 

Example 5. — Initial velocity, 60 feet per second ; mass weighs 
0.30 of a ton. A resistance of 112f lbs. retards it f or -£% of 
a minute. Required the distance passed over during this time. 

Ans. 286.8 feet. 

Example 6. — Required the time in which a force of 600 avoir- 
dupois ounces will increase the velocity of a mass weighing 1-J- 
tons from 480 feet per minute to 240 inches per second. 

Ans. 30 seconds. 

Example 7. — What distance is passed over by a mass of (0.6) 
tons weight during the overcoming of a constant resistance 
(friction), if its velocity, initially 144 inches per sec, is reduced 
to zero in 8 seconds. Required, also, the friction. 

Ans. 48 ft. and 55 lbs. 

Example 8. — Before the action of a force (value required) a 
body of 11 tons had described uniformly 950 ft. in 12 minutes. 
Afterwards it describes 1650 feet uniformly in 180 seconds. 
The force acts 30 seconds. P — ? Ans. P = 178 lbs. 



RECTILINEAR MOTION OF A MATERIAL POINT. 57 

58. Graphic Representations. Unif. Ace. Motion. — With the 

initial velocity = 0, the equations of § 56 become 

v=p, (1) • = **?, (2) 

s = v* -=- 2j>, . . . (3) and s = ivt (4) 

Eqs. (1), (2), and (3) contain each two variables, which may 
graphically be laid off to scale as co-ordinates and thus give a 
curve corresponding to the equation. Thus, Fig. 62, in (I.), we 



A 

V 

!T 


s 


*■ —/ • 


MQ---^--- 7 / 




u 


(II.) 
Fig. 62. 


(III.) 



have a right line representing eq. (I.) ; in (II.), a parabola with 
axis parallel to *, and vertex at the origin for eq. (2) ; also a 
parabola similarly situated for eq. (3). Eq. (4) contains three 
variables, s, v, and t. This relation can be shown in (I.), s be- 
ing represented by the area of the shaded triangle = \vt. 
(II.) and (III.) have this advantage, that the axis OS may be 
made the actual path of the body. [Let the student determine 
how the origin shall be moved in each case to meet the supposi- 
tion of an initial velocity == -f- c or — <?.] 

59. Variably Accelerated Motions. — We here restate the equa- 
tions 

v = di . . <L);p = -£ = ■£. . (JL);vdv=pds..(IIL); 

and resultant force 

= P=Mp, (IV.); 

which are the only ones for general use in rectilinear motion. 

Problem 1. — In pulling a mass M along a smooth, horizon- 
tal table, by a horizontal cord, the tension is so varied that 
s = If (not a homogeneous equation / the units are, say, the 
foot and second). Required by what law the tension varies. 



58 MECHANICS OF ENGINEERING. 

From (I.) v = dt = -^ = ISf ; from (II.), p = -^ = 

24tt; and (IY.) the tension — P — Mp = 2±Mt, i.e., varies 
directly as the time. 

Problem 2. " Harmonic Motion," Fig. 63. — A small block 



-M 

jo 



-s Q.....^ z -^-*P +s 

Fig. 63 

on a smooth horizontal table is attached to two horizontal 
elastic cords (and they to pegs) in such a way that when the 
block is at 0, each cord is straight but not tense ; in any other 
position, as m, one cord is tense, the other slack. The cords 
are alike in every respect, and, as with springs, the tension 
varies directly with the elongation (= s in figure). If for an 
elongation s, the tension is T x , then for any elongation s it is 
T = T,s -T- s,. The acceleration at any point m, then, is 
p — — (T -T- M) = — (2> -T- Ms,), which for brevity put 
p = — as, a being a constant. Required the equations of 
motion, the initial velocity being = -\- c, at 0. From eq. (III.) 

vdv = — a I sds* 

t/O 

i.e., ^{f — c 2 ) = — ias* ; or, v 2 = c 2 — as 2 . . (1) 
From (I.), dt = ds -T- v; hence from (1), 

£clt = £[_ds - V7~~^T S \ 



or 



V^e/o V 1 _( si / a _^ c y VaU \ c J 



-psm- 1 ( ; (2) 

Va v o I v J 



RECTILINEAR MOTION OF A MATERIAL POINT. 59 



Inverting (2), we have s == (e -~ Va) sin (t Va), . . . (3) 
Again, by differentiating (3), see (I.), v = c cos (t Va) (4) 
Differentiating (4), see (H.),p = — c Va sin (t Va). . . (5) 

These are the relations required, but the peculiar property 
of the motion is made apparent by inquiring the time of pass- 
ing from to a state of rest ; i.e., put v = in equation (4), 
we obtain t = \rt -=- Va, or f tt -f- Va, or J-7T 4- Va, and so on, 
while the corresponding values of s (from equation (3))« are 
+ (c 4- Va), — (c -T- Va), -f- (c -f- Va), and so on. This shows 
that the body vibrates equally on both sides of in a cycle or 
period whose duration = 2?r 4- Va, and is independent of the 
initial velocity given it at 0. Each time it passes O the 
velocity is either -\- c, or — <?, the acceleration = 0, and the 
time since the start is = nn 4- Va, in which ^ is any whole 
number. At the extreme point p = =F c Va, from eq. (5). 
If then a different amplitude be given to the oscillation by 
changing c, the duration of the period is still the same, i.e., 
the vibration is isochronal. The motion of an ordinary pen- 
dulum is nearly, that of a cycloidal pendulum exactly, harmonic. 

If the crank-pin of a reciprocating engine moved uniformly 
in its circular path, the piston would have a harmonic motion 
if the connecting-rod were infinitely long, or if the design in; 



•< — 2r- 



JILJII 



m m ~c 




Fig. 64. 

Fig. 64 were used. (Let the student prove this from eq. (3).) 
Let 2r = length of stroke, and c = the uniform velocity of the 
crank-pin, and M = mass of the piston and rod AJB. Then 
the velocity of M at mid-stroke must = c, at the dead-points, 
zero ; its acceleration at mid-stroke zero ; at the dead-points 
the ace. = c Va, and s = r = c -f- Va (from eq. (3)) ; .*. Va 
= e -T- r, and the ace. at a dead point (the maximum ace.} 



MECHANICS OF ENGINEERING. 



= & -T- r. Hence on account of the acceleration (or retarda- 
tion) of M in the neighborhood of a dead-point a pressure will 
be exerted on the crank-pin, equal to mass X ace. = Mb* -4- r 
at those points, independently of the force transmitted due to 
steam-pressure on the piston-head, and makes the resultant 
pressure on the pin at G smaller, and at D larger than it would 
be if the "inertia" of the piston and rod were not thus taken 
into account. We ma}' prove this also by the free-body method, 
considering AB free immediately after passing the dead-point 

G, neglecting all friction. See Fig. 



v 



J9 



Fig. 65. 



P' 



65. 



The forces acting are : G, 



the 
weight ; J¥, the pressures of the 
guides ; P, the known effective steam- 
pressure on piston-head ; and JP', the unknown pressure of 
crank-pin on side of slot. There is no change of motion ver- 
tically ; .-. N' -f N— G = 0, and the resultant force is P — P' 
== mass X accel. = Me 2 -f- r, hence P' = P — M& -=- r. 
Similarly at the other dead-point we would obtain P' = P -f- 
Mc* -f- r. In high-speed engines with heavy pistons, etc., 
Me 2 -T- r is no small item. [The upper half-revol., alone, is 
here considered.] 

Problem 3. — Supposing the earth at rest and the resistance 
of the air to he null, a body is given an initial upward vertical 
velocity = o. Required the velocity at any distance s from 
the centre of the earth, whose attraction varies in- 
versely as the square of the distance s. 

See Fig. 66. — The attraction on the body at the 
surface of the earth where s = r, the radius, is its 
weight G ; at any point m it will be P = G(r 2 -=- s 2 ), 
while its mass = G -+- g. 

Hence the acceleration at m =p = ( — P) -^ M 




Fig. 66. 



= - ff(r' h- «*). 
and we have 

vdv = - 



Take equation III., vdv = _pds, 



gr*s ~ 2 ds; 



vdv = — gr* J s~ 2 ds; or, %v 2 = — gr 
i.e., i(V - c 2 ) = - gr' f- - 



PI 

L s> 



(i) 



RECTILINEAR MOTION OF A MATERIAL POINT. 



61 



Evidently v decreases, as it should. Now inquire how small 
a value c may have that the body shall never return/ i.e. r 
that v shall not = until s = oo. Put v = and s = oo in 
(1) and solve for c ; and we have 



o = Vtyr = V2 X 32.2 X 21000000, 

= about 36800 ft. per sec. or nearly 7 miles per sec. Con- 
versely, if a body be allowed to fall, from rest, toward the 
earth, the velocity with which it would strike the surface 
would be less than seven miles per second through whatever 
distance it may have fallen. 

If a body were allowed to fall through a straight opening in 
the earth passing through the centre, the motion would be har- 
monic, since the attraction and consequent acceleration now 
vary directly with the distance from the centre. See Prob. 2. 
This supposes the earth homogeneous. 

Problem 4. — Steam working expansively and raising a weight. 



-fl G| 




Fig. 67. 



Fig. 67. — A piston works without 
friction in a vertical cylinder. Let 
S = total steam-pressure on the 
underside of the piston ; the weight 
G, of the mass G -f- g (which in- 
cludes the piston itself) and an 
atmospheric pressure = A, con- 
stitute a constant back-pressure. 
Through the portion OB = * 1S of 
the stroke, S is constant = S 19 while beyond B, boiler com- 
munication being " cut off," S diminishes with Boyle's law, i.e., 
in this case, for any point above B, we have, neglecting the 
"clearance", F being the cross-section of the cylinder, 

S\S x \\Fs x \Fs\ or 8=S jSl -±-s. 

Pull length of stroke = ON = s n . Given, then, the force* 
S x and A, the distances s t and s w and the velocities at O and 
at N both = (i.e., the mass M = G -+- g is to start from rest at 
O, and to come to rest at iV), required the proper weight G to 



62 MECHANICS OF ENGINEERING-. 

fulfil these conditions, 8 varying as already stated. The accel- 
eration at any point will be 

p=\_S-A-G] + M. . . . . (1) 

Hence (eq. III.) Mvdv = [8 — A — G]ds, and .*. for the 
whole stroke 

M£vdv=f o [S — A.-r- G]ds; i.e., 

/si r s *ds p s * p Sa 

ds + 8,sJ s — - Aj^ ds- GJ o ds, 



or 8 A [l + log, j] = As n + Gs n . . . . (2) 

Since 8 = 8 1 = constant, from to B, and variable, = 
8 l s 1 -~ 8, from B to iV, we have had to write the summation 

x 



N 

Sds in two parts. 



From (2), G becomes known, and .*. M also (= G ■— g). 
Required, further, the time occupied in this upward stroke. 
From to B (the point of cut-off) the motion is uniformly 
accelerated, since p is constant (8 being == 8 1 in eq. (1) ), 
with the initial velocity zero; hence, from eq. (3), § 56, 
the velocity at B = v 1 = V2 [8, — A — G]s x -=- M is known ; 
.*. the time t x == 2s x -i- v 1 becomes known (eq. (4), § 56) of de- 
scribing OB. At any point beyond B the velocity v may be ob- 
tained thus : From (III.) vdv = pds, and eq. (1) we have, 
summing between B and any point above, 

Mf\dv = S A £ j-(A + G)£ds ; i.e., 

| (^_-jO = Sa log. ^-(A + G) (s - s,). 

This gives the relation between the two variables v and 8 
anywhere between B and iV; if we solve for v and insert its 
value in dt = ds -r- v, we shall have dt = a function of s and 
ds, which is not integrable. Hence we may resort to approxi- 



KECTILINEAR MOTION OF A MATERIAL POINT. 63 

mate methods for the time from B to N. Divide the space 
^iTinto an uneven number of equal parts, say five; the dis- 
tances of the points of division from will be s v <s 2 , s 3 , s 4 , s b , 
and s n . For these values of s compute (from above equation) 
v x (already known), v„ v„ v 4 , v M and v n (known to be zero). To 
the first four spaces apply Simpson's Rule, and we have the 
time from B to the end of s b , 

[_t =J i -5 approx. = -^- ^- + - + - +- +-J; 

while regarding the motion from 5 to N~ as uniformly retarded 
(approximately) with initial velocity = v b and the final = zero, 
we have (eq. (4), § 56), 
-n 

t = 2(s n - s 5 ) -i- v b . 

-6 

By adding the three times now found we have the whole time 
of ascent. In Fig. 67 the dotted curve on the left shows by 
horizontal ordinates the variation in the velocity as the distance 
s increases ; similarly on the right are ordinates showing the 
variation of S. The point E, where the velocity is a maximum 
== v m , may be found by putting p = 0, i.e., for S = A -f- G, 
the acelerating force being = 0, see eq. (1). Below .2^ the ac- 
celerating force, and consequently the acceleration, is positive ; 
above, negative (i.e., the back-pressure exceeds the steam- 
pressure). The horizontal ordinates between the line HE' KL 
and the right line RT&yq proportional to the accelerating force. 
If by condensation of the steam a vacuum is produced be- 
low the piston on its arrival at 2V", the accelerating force is 
downward and = A -f- G. [Let the student determine how 
the detail of this problem would be changed, if the cylinder 
were horizontal instead of vertical.] 

60. Direct Central Impact. — Suppose two masses M x and J/ 2 
to be moving in the same right line so that their distance apart 
continually diminishes, and that when the collision or impact 
takes place the line of action of the mutual pressure coincides 
with the line joining their centres of gravity, or centres of 



64 MECHANICS OF ENGINEERING. * 

mass, as they may be called in this connection. This is called 
a direct central impact, and the motion of each mass is varia- 
bly accelerated and rectilinear during their contact, the only 
force being the pressure of the other body. The whole mass 
of each body will be considered concentrated in the centre of 
mass, on the supposition that all its particles undergo simul- 
taneously the same change of motion in parallel directions. 
(This is not strictly true ; the effect of the pressure being 
gradually felt, and transmitted in vibrations. These vibrations 
endure to some extent after the impact.) When the centres 
of mass cease to approach each other the pressure between the 
bodies is a maximum and the bodies have a common velocity ; 
after this, if any capacity for restitution of form (elasticity) 
exists in either bod} 7 , the pressure still continues, but dimin- 
ishes in value gradually to zero, when contact ceases and the 
bodies separate with different velocities. Reckoning the time 
from the first instant of contact, let t' = duration of the first 
period, just mentioned ; t" that of the first -|- the second (resti- 
tution). Fig. 68. Let M x and M^ be the masses, and at any 
instant during the contact let v x and v 9 . 
be simultaneous values of the velocities 
of the mass-centres respectively (reckon- 
FlG - 68, ing velocities positive toward the right), 

and P the pressure (variable). At any instant the acceleration 
of M x is p x = — (P -r" -3/,), while at the same instant that of 
M 2 is j? 2 = -\- (P -±- M 2 ) ; M x being retarded, M 2 accelerated, 
in velocity. Hence (eq. II., p = dv -=- dt) we have 

M x dv x = - Pdt; and M s dv t = + Pdt. . . (1) 

■Summing all similar terms for the first period of the impact,. 
we have (calling the velocities before impact c x and <? 2 , and the 
common velocity at instant of maximum pressure C) 

■#"jft= - •/***> Le - M >(° -^ = - S! Pdt ; ( 2 > 

M, fflv, = +f i '-Pdt, i.e., M,{C - a,) = +fPJt. . (3> 



r p< 


vH 


-^ 


- 1 < 


11, 1 




v* 


1 p 




M, 


M 2 





RECTILINEAR MOTION OF A MATERIAL POINT. 65 

The two integrals are identical, numerically, term by term, 
since the pressure which at any instant accelerates J/ 2 is nu- 
merically equal to that which retards M^\ hence, though we do 
not know how P varies with the time, we can eliminate the 
definite integral between (2) and (3) and solve for 0. If 
the impact is inelastic (i.e., no power of restitution in either 
body, either on account of their total inelasticity or damaging 
effect of the pressure at the surfaces of contact), they continue 
to move with this common velocity, which is therefore their 
final velocity. Solving, we have 

„_ M& + M& 

°~ M,+M 2 W 

Next, supposing that the impact is partially elastic, that the 
bodies are of the same material, and that the summation 

/ Pdt for the second period of the impact bears a ratio, e. 

to that / Pdt, already used, a ratio peculiar to the material, 

if the impact is not too severe, we have, summing equations 
(1) for the second period (letting Y x and V 2 = the velocities 
after impact), 

M, f c Vl dv x = - f t ?Pdt, i.e., MX V- C) = - eJ*Pdt ; (5) 
M t f**d» % = + fJ'Pdt, i.e., Ml V-0) = + e f Pdt. (6) 

«/ 

e is called the coefficient of restitution. 

Having determined the value of / Pdt from (2) and (3) in 

terms of the masses and initial velocities, substitute it and that 
of O, from (4), in (5), and we have (for the final velocities) 

V v = [Jf A + Mj % - eM^-cj] - [Jf, + JfJ; . (7) 
and similarly 

V^lMA + M^+eMfa-cM+lMt+Mj. . (8) 
For e = 0, i.e., for inelastic impact, V z = V^= C in eq. (4) ; for 



66 MECHANICS OF ENGINEERING. 

t ±= 1, or elastic impact, (7) and (8) become somewhat simpli- 
fied. 

To determine e experimentally, let a ball (Jf,) of the sub- 
stance fall upon a very large slab (JfQ of the same substance, 
noting both the height of fall A 1} and the height of rebound H x . 
Considering Jf 2 as = oc, with 



c x — V 2gh„ V t = — V 2gJS' iy and c 2 = o, 

eq. (7) gives 



-V2gH 1 ,=-eV2gh 1 ', .\e = VH x + h x . 

Let the student prove the following from equations (2), (3), 
(5), and (6) : 

(a) For any direct central impact whatever, 

Jf A + M,c, = M x V x + M % V v 

[The product of a mass by its velocity being sometimes 
called its momentum, this result may be stated thus: 

In any direct central impact the sum of the momenta before 
impact is equal to that after impact (or at any instant during 
impact). This principle is called the Conservation of Momen- 
tum. The present is only a particular case of a more general 
proposition. 

It may be proved C, eq. (4), is the velocity of the centre of 
gravity of the two masses before impact ; the conservation of 
momentum, then, asserts that this velocity is unchanged by the 
impact, i.e., by the mutual actions of the two bodies.] 

(J>) The loss of velocity of M x , and the gain of velocity of 
M» are twice as great when the impact is elastic as when in- 
elastic. 

(c) If e = 1, and M 1 = M n then Y x — + c 3 , and F 2 = e lm 

Example. — Let M x and M 2 be perfectly elastic, having weights = 4 and 
5 lbs. respectively, and let c x = 10 ft. per sec. and c 2 = — 6 ft. per sec. 
(i. e., before impact M 2 is moving in a direction contrary to that of M^). 
By substituting in eqs. (7) and (8), with e = 1, Mi = 4 -*- g, and M 2 = 5 -r- g, 
we have 

7i = I|~4 x 10 + 5 x (- 6) - 5 (10 - (- 6 ))]= ~ 7 - 7 ft P er sec * 

F 2 = I ["4 x 10 + 5 x (~ 6) + 4 flO - (- 6))J= + 8.2 ft. per sec. 

as the velocities after impact. Notice their directions, as indicated by their 



u 



VIRTUAL VELOCITIES." 67 



CHAPTER II. 

"VIRTUAL VELOCITIES." 

61. Definitions.— If a material point is moving in a direction 
not coincident with that of the resultant force acting (as in 
curvilinear motion in the next chapter), and any element of its 
path, ds, projected upon this force;* the length of this projee- 
tion, du, Fig. 69, is called the "Viktual Velocity" of the 
torce, since du ~ dt may be considered the veloc- 
ity of the force at this instant, just as ds -=- dt is 7 /-"P 
that of the point. The product of the force by &*f 
its du will be called its virtual moment, reckoned ^^S^ 
+ or — according as the direction from O to D is FlG - 69 - 
the same as that of the force or opposite. 

62. Prop. L-Tfo virtual moment of a force equals the 
algebraic sum of those of its components. Fig. 70. Take the 

P . direction of ds as an axis X; let P and P 

■P be components of P- „„ „ o and ' a tllei ; 
angles with X. Then (§ 16) P cos a = 
Poos a l+ P t cos ^ Hence p(& cog . 

^ P,(ds cos a,) + P^ds cos a,). But ds cos a 

H. 70 ' 7 the l 3ro J ectio " of * UP"" P, La, = <fe ; 

^ If in Fig. 70 „, were > 90°, evidently we would 
have Pdu = -.p idui + p tdu U >& J ou ^ 

negative, and ^, would fall behind Whence the deflnidon 

Li a i7/ n § }■ FW ' a ' ly nnmber of components the 
p oof would be similar, and is equally applicable whether they 
aie in one plane or not. J 

63. Prop. II The sum of the virtual moments equals zero, 
f or concurrent forc es in equilibrium, 

b^tZTr^ mt ^ I8&7 "P ro J ecte donthe line of action ofl^W^ 

irat the phrase used may be allowed, for brevity. ' 



68 MECHANICS OF ENGINEERING. 

(If the forces are balanced, the material point is moving in 
a straight line if moving at all.) The resultant force is zero. 
Hence, from § 62, P x du x + PA + etc. = 0, having proper 
regard to sign, i.e., 2(Pdu) = 0. 

64. Prop. III.— The sum of the virtual moments equals zera 
for any small displacement or motion of a rigid body in equi- 
librium under non-concurrent forces in a plane; all points of 
the body moving parallel to this plane. (Although the kinds 
of motion of a given rigid body which are consistent with 
balanced non-concurrent forces have not yet been investigated, 
we may imagine any slight motion for the sake of the alge- 
braic relations between the different du's and forces.) 

First, let the motion be a translation, all 
^_/ points of the body describing equal parallel 

\//TV lengths = ds. Take X parallel to ds ; let a x , 
f\ dsds ^ \5retc, be the angles of the forces with X 
I V '' -pf Then(§35)5(Pcosa) = 0;.-.^^(Pcosar) 
\ •--. y* = ; but ds cos a x = du x ; ds cos « 2 = du % ; 
M^y etc. ; /. 2{Pdu) = 0. Q. E. D. 
I - Secondly, let the motion be a rotation 

fig. 71. through a small angle dd in the plane of the 

forces about any point in that plane, Fig. 72. With as apole 
let p x be the radius-vector of the point of application of P„ and 
a, its lever-arm from 0\ similarly for the 
other forces. In the rotation each point of 
application describes a small arc, ds» ds„ 
etc., proportional to p„ p a , etc., since ds, 
*=.p x dd, ds, = p t dd, etc. Fiw § 36,^;^. 
Pjt x _|_ etc. = ; but from similar triangles ""'•••-a r .^ / 
ds\ : du x : : Pl : a x ; >. a x = P.duj ds x ^ 

r= du x S- dd ; similarly a t = du a -r- dV, etc. 
Pence we must have \_P x du x + PA + • • •] ■*" dd = °> La » 
2(P^) = 0. Q. E. D. 

Now since any slight displacement or motion of a body may 
be conceived to be accomplished by a small translation fol- 
lowed by a rotation through a small angle, and since the fore- 




going deals only with, projections of paths, the proposition is 
established and is called the Principle of Virtual Velocities. 
[A similar proof may be used for any slight motion what- 
ever in space when a system of non-concurrent forces is bal- 
anced.] Evidently if the path (ds) of a point of application is 
perpendicular to the force, the virtual velocity {die), and con- 
sequently the virtual moment (Pdu) of the force are zero. 
Hence we may frequently make the displacement of such a 
character in a problem that one or more of the forces may be 
excluded from the summation of virtual moments. 

65. Connecting-Rod by Virtual Velocities. — Let the effective 
steam-pressure P be the means, through the connecting-rod 
and crank (i.e., two links), of raising the weight G very slowly; 
neglect friction and the weight of the links themselves. Con- 
sider AB as free (see (o) in Fig. 73), BO also, at (c); let the 

kdx% ^ C 



"small displacements" of both be simultaneous small portions 
of their ordinary motion in the apparatus. A has moved to A 1 
through clx ; B to B^ through ds, a small arc ; C has not 
moved. The forces acting on AB are P (steam-pressure), N 
(vertical reaction of guide), and N' and T(the tangential and 
normal components of the crank-pin pressure). Those on BC 
are N' and T (reversed), the weight G, and the oblique pressure 
of bearing P '. The motion being slow (or rather the accelera- 
tion being small), each of these two systems will be considered as 
balanced. Now put ^(Pdi() = for AB, and we have 

Pdx + IT X + IT X - Tds = 0. . . (1) 

For the simultaneous and corresponding motion of BC, 
2(Pdu) = gives 



ro 



MECHANICS OF ENGINEERING. 

JV X + Tds - Gdh + P' x = 0, 



(2) 



dh being the vertical projection of 6r's motion. 

From (1) and (2) we have, easily, Pdx — Gdh = 0, . (3) 

./ B i which is the same as we might have 

\ ^''B^Q^\~ h obtained by pntting ^2{Pdu) = for 

■^*a"""""£ '' ' ' ' ' gScP' ^ e ^ wo ^' in ^ s together, regarded col- 
* \ lectively as a free body, and describ- 
Flcf - 74 - ing a small portion of the motion 

they really have in the mechanism, viz., (Fig. 74,) 

Pdx + lfxO- Gdh + P' X = 0. . . (4) 

We may therefore announce the — 



66. Generality of the Principle of Virtual Velocities. — If any 

mechanism of flexible inextensible cords, or of rigid bodies 
jointed together, or both, at rest, or in motion with very small 
accelerations, be considered free collectively {or any portion of 
it), and all the external forces ptit in ; then {disregarding 
mutual frictions) for a small portion of its prescribed motion, 
2(Pdu) must = 0, in which the du, or virtual velocity, of 
each force, P, is the projection of the path of the point of 
application upon the force (the product, Pdu, being -\- or — 
according to § 61). 

67. Example. — In the problem of § 65, having given the 
weight G, required the proper steam-pressure (effective) P to 
hold G in equilibrium, or to raise it uniformly, if already in 
motion, for a given position of the links. That is. Fig. 75, 
given a, r, c, a, and (3, re- 
quired the ratio dh : dx ; for, 
from equation (3), § 65, P 
= G(dh : dx). The projec- 
tions of dx and ds upon AB 
will be equal, since AB = 
A^B^ and mnkes an (infinitely) small angle with A^B^ i.e., 
dx cos a — ds cos (ft — a). Also, dh = (c : ?*)ds sin ft. 




dx A, 



Fig. 75. 



71 

Eliminating ds, we have, 

dh c sin ft cos a c sin (3 cos a 

dx ~ r cos (ft — a) ' ' ' ' ~ /* cos (/? — a)* 



68. "When the acceleration of the parts of the mechanism is 
not practically zero, 2{Pdu) will not = 0, but a function of 
the masses and velocities to be explained in the chapter on 
"Work, Energy, and Power. If friction occurs at moving joints, 
enough " free bodies" should be considered that no free body 
extend beyond such a joint ; it will be found that this friction 
cannot be eliminated in the way in which T and N' were, in 
§65. 

69. Additional Problems ; to be solved by " virtual velocities." 
Problem 1. — Find relations between the forces acting on a 
straight lever in equilibrium ; also, on a bent lever. 

Problem 2. — When an ordinary copying-press is in equilib- 
rium, find the relation between the force applied horizontally 
and tangentially at the circumference of the wheel, and the 
vertical resistance under the screw-shaft. 

Solution.— Considering free the rigid body consisting of the wheel and 
screw-shaft, let R be the resistance at the point of the shaft (pointing 
along the axis of the shaft), and P the required horizontal tangential force 
at edge of wheel. Let radius of wheel be r. Besides R and P there are 
also acting on this body certain pressures, or " supporting forces,'' consist- 
ing of the reactions of the collars, and reactions of the threads of nut against 
the threads of screw. Denote by s the "pitch " of the screw, i.e., the dis- 
tance the shaft would advance for a full turn of the wheel. Then if we 
imagine the wheel to turn through a small angle dd, the corresponding 

advance, ds, of the shaft would be — , from the proportion s : ds ::27t : dQ. 

The path of the point of application of P would be a small portion of 
a helix, the projection of which on the line of P is rdB, while ds projects 
in its full length on the line of the force R. In the case of each of the 
other forces, however, the path of the point of application is perpendicular 
to the line of the force (which is normal to the rubbing surfaces, friction 
being disregarded). Hence, substituting in ~E(Pdu) = 0, we have 

+ P.rdQ-R. ds + + = 0; 
whence 

r ~ rdh- U -^nr' K ' 



72 



MECHANICS OF ENGINEERING. 



CHAPTER III. 



CURVILINEAR MOTION OF A MATERIAL POINT. 



A° x ° 

o / 


D° 


c nil """"■; 




o A// B/ 


Jim 


{ JJ_ | |^| - 



Fig. 76. 



[Motion in a plane, only, will be considered in this chapter.] 

70. Parallelogram of Motions. — It is convenient to regard 
the curvilinear motion of a point in a plane as compounded, or 
made up, of two independent rectilinear motions parallel 
respectively to two co-ordinate axes X and Y, as may be ex- 
plained thus : Fig. 76. Consider the 
drawing-board CD as fixed, and let the 
head of a T-square move from A 
toward B along the edge according to 
any law whatever, while a pencil moves 
from M toward Q along the blade. The 
result is a curved line on the board, whose 
form depends on the character of the 

two JTand Y component motions, as they may be called. If 
m a time t x the T-square head has moved an ^distance = MB, 
and the pencil simultaneously a Y distance = MP, by com- 
pleting the parallelogram on these lines, we obtain i?, the 
position of the point on the board at the end of the time t x . 
Similarly, at the end of the time t/ we find the point at JR f . 

71. Parallelogram of Velocities. — Let the X and Amotions 
be uniform, required the resulting motion. Fig. 77. Let c^ 
and c y be the constant uniform JTand ^velocities. Then in 
any time, t y we have x = c x t and y = 

we have, eliminating t, 
- c y = constant, i.e., x is 

the path is a ( 

off OA = cl. 



c y t ; whence 
x 



y = *■ 

proportional to y, 
straight line. Laying 



i.e., 




and A B = c y , B is a point of the path, 

and OB is the distance described by the point in the first 



CURVILINEAR MOTION OF A MATERIAL POINT. 73 



second. Since by similar triangles OR : x : : OB : o m we 
Lave also OR = OB . t ; hence the resultant motion is uniform, 
and its velocity, OB = c, *s the diagonal of the parallelogram 
formed on the two component velocities. 

Corollary. — If the resultant motion is curved, the direction 
and velocity of the motion at any point will be given by the 
diagonal formed on the component velocities at that instant. 
The direction of motion is, of course, a tangent to the curve. 

72. Uniformly Accelerated X and Y Motions. — The initial 
velocities of both being zero. Required the resultant motion. 
Fig. 78. From § 56, eq. (2) (botli c^ andc^ y / , 
being = 0), we have x = \p x t and y = ft ^^/ 
ip y f, whence x -i- y = jp x -r- Py— constant, WvZ^J / 

and the resultant motion is in a straight tL—-d®& L — x 

line. Conceive lines laid off from on X °* x 

and Y to represent^ and_p v to scale, and FlG " 78# 

form a parallelogram on them. From similar triangles {OR 
being the distance described in the resultant motion in any 
time t), OR : x : : ~OB : p x ; .-. Oif— \OBi\ Hence, from the 
form of this last equation, the resultant motion is uniformly 
accelerated, and its acceleration is OB =- p, (on the same scale 
a&p x 2LT\dp y ). 

This might be called the parallelogram of accelerations, but 
is really a parallelogram of forces, if we consider that a free 
material point, initially at rest at 0, and acted on simulta- 
neously by constant forces P x and P y (so that p x = P x -^ M 
and^, = P y -r- M\ must begin a uniformly accelerated recti- 
linear motion in the direction of the resultant force, having no 
initial velocity in any direction. 

73. In general, considering the point hitherto spoken of as a 
free material point, under the action of one or more forces, in 
view of the foregoing, and of Newton's second law, given the 
initial velocity in amount and direction, the starting-point, 
the initial amounts and directions of the acting forces and the 



74 



MECHANICS OF ENGINEERING. 



laws of their variation if they are not constant, we can resolve 
them into a single X and a single Y force at any instant, 
determine the X and Amotions independently, and afterwards 
the resultant motion. The resultant force is never in the direc- 
tion of the tangent to the path (except at a point of inflection). 
The relations which its amount and direction at any instant 
bear to the velocity, the rate of change of that velocity, and 
the radius of curvature of the path will appear in the next 
paragraph. 



74. General Equations for the curvilinear motion of a ma- 
terial point in a plane. — The motion will be considered result- 



'-\l 



M' 







Fig. 



ing from the composition of 
independent Xand Y motions, 
X m\d Y being perpendicular to 
each other. Fig. 79. In two 
consecutive equal times (each 
= dt), let dx and dx' = small 
spaces due to the X motion ;, 
and dy and CK ' = dy', due to* 
the Y motion. Then ds and 
ds' are two consecutive elements 
of the curvilinear motion. Pro- 
long ds. making BE=ds; then EE '= d*x, CE=d*y, and 
CO = d 2 s (EO being perpendicular to BE). Also draw CL 
perpendicular to BG and call CL d 2 n. Call the velocity of 
the X motion v x , its acceleration p x \ those of the Amotion, 
v y and p y . Then, 

dx dy dv x d*x , _ dv^ 

* x=z dT> Vy = ~dt> Px = ~dt = dtf> ^ Pv = ~dt 

For the velocity along the curve (i.e., tangent) 
1) r= ds -7- dt, we shall have, since ds* = dx* -f- dy*, 

dt) ' '' [diJ "•" \cti) ~ 



+ <• 



9y_ 

df 



(1) 



Hence v is the diagonal formed on v x and v y (as in § 71). 
Let p t = the acceleration of v, i.e., the tangential acceleration, 



CURVILINEAR MOTION OF A MATERIAL POINT. 75 



then jp t = d % s -— df, arid, since d 2 s = the sum of the projec- 
tions of EF 'and CF on BC, i.e., d 2 s = <$ a a? cos « -|- d*y sin #> 
we have 

d 2 s d 2 x . d 2 y . . , 

df = df G0S a ^"df Sm " ' 1 * e * 5 ^ f = i?iC C0S " +^ Sin "" ' ' 

By Normal Acceleration we mean the rate of change of the 
velocity in the direction of the normal. In describing the ele- 
ment AB = ds, no progress has been made in the direction of 
the normal BIT i.e., there is no velocity in the direction of the 
normal; but in describing BG (on account of the new direc- 
tion of path) the point has progressed a distance CL (call it 
d?ri) in the direction of the old normal BH (though none in 
that of the new normal CI). Hence, just as the tang. ace. 



ds' — ds 



= -t^t, so the normal accel. = 



CL 



zero d 2 n 



~~ df ~df'^ * " df 

It now remains to express this normal acceleration (=p n ) in 
terms of the X and Y accelerations. From the figure, CL 
= CM- ML, i.e., 

d 2 n = d 2 y cos a — d 2 x sin a {since EF = d 2 x\ ; 



dt 2 



d 2 y 

= df C0Sa ~ 



d 2 x 



sm a. 



df 

Hence JPn^Jpy cos a — i^sin a (3) 

The norm. ace. may also be expressed in terms of the tang, 
velocity v, and the radius of curvature r, as follows : 
ds' = rda, or da = ds' -f- r ; also d 2 n = ds'da, = ds' 2 -~ r, 
d 2 n (ds'Vl v 2 

dt=\di) P ° r *« = r (*) 

80, we resolve the forces X = Mp x and 7 
= Mp y , which at this instant account for the 
X and Y accelerations (M = mass of the 
material point), into components along the 
tangent and normal to the curved path, we 
shall have, as their equivalent, a tangential 
force 

T = Mp x cos a -f- Mjp y sin <*, 



i.e., 



If now, Fig. 

Y 

.--ft. 



v 



>#*■ 



<»J 



*©- 



<2f 



*X 



Fig. 



76 MECHANICS OF ENGINEERING. 

and a normal force 

JV = Mp y cos a — Mp x sin a. 
But [see equations (2), (3), and (4)] we may also write 

T=Mp t = M T( ; and W=Mp n = M-. . (5) 

Hence, if a free material point is moving in a curved path, 
the sum of the tangential components of the acting forces must 
equal (the mass) X tang, accel.; that of the normal components, 
= (the mass) X normal accel. = (mass) X (square of veloc. in 
path) -f- (rad. curv.). 

It is evident, therefore, that the resultant force (= diagonal 
on T and N or on JTand Y, Fig. 80) does not act along the tan- 
gent at any point, but toward the concave side of the path ; un- 
less r = oo. 

Radius of curvature. — From the line above eq. (4) we 
have d 2 n = ds' 2 -~ r ; hence (line above eq. (3) ), ds n -^ r — 
d 2 y cos a — d*x sin a ; but cos a = dx^- ds, and sin a — dy-^ ds, 

ds' 2 ,„ dx ^ dy ds' 2 ds 7 Tdxd*y — dyd*x~] 

•■-T=< p yTs-< Fay £> <»• — =<H — & — J ; 

i.e., — dx 2 d\ -Jf- — dx 2 d (tan a), 

_ fds' 2 ds\ r/ dx\ 2 d tan an 
"'" r ~ \W~) * \_\dtJ ~~~dT\ ' 

, = ,. + [„,^-] « 

which is equally true if, for v x and tan a, we put v y and 

tan (90° — a ), respectively. 

Change in the velocity square. — Since the tangential accelera- 

dv dv _.'■■•. 

tion -7- =J9j, we nave ^^ =p t as; i.e., 

j-dv=p t ds, or vdv =p t ds and .*. — - — = / p t ds. (7) 

having integrated between any initial point of the curve where 
w = c, and any other point where v = v. This is nothing 
more than equation (III.), of § 50. 




CURVILINEAR MOTION OF A MATERIAL POINT. 77 

75. Normal Acceleration. Second Method. — Fig. 81. Let 
C be the centre of curvature and OD = 2r. Let OB' be a 
portion of the oscillatory parabola (vertex at D v N 
O ; any osculatory curve will serve). When 
ds is described, the distance passed over in 
the direction of the normal is AB ; for 2ds, 
it would be A'B' = ±AB (i.e., as the 
6quare of OB'; property of a parabola), 
and so on. Hence the motion along the normal is uniformly 
accelerated with initial velocity = 0, since the distance AB, 
varies as the square of the time (considering the motion along 
the curve of uniform velocity, so that the distance OB is di- 
rectly as the time). If jp n denote the accel. of this uniformly 
accelerated motion, its initial velocity being = 0, we have (eq. 
2, § 56) AB =± %j> n dt% i.e.,^ n = %Ab ~ dt\ _But from the 
similar triangles ODB and OAB we have, AB : dsi'.ds : 2r, 
hence 2AB = ds 2 -r-r, .'.p n = ds 2 -=- rdtf = v 2 -=- r. 

76. Uniform Circular Motion. Centripetal Force. — The ve- 
locity being constant, ^ must be = 0, and .*. 2\ov ^Tif there 
are several forces) must = 0. The resultant of all the forces, 
therefore, must be a normal force = (Mc 2 -f- r) = a con- 
stant (eq. 5, § 74). This is called the " deviating force," 
or " centripetal force ;" without it the body would continue 
in a straight line. Since forces always occur in pairs (§ 3), 
a " centrifugal force," equal and opposite to the " centri- 
petal" (one being the reaction of the other), will be found 
among the forces acting on the body to whose constraint the 
deviation of the first body from its natural straight course is 
due. For example, the attraction of the earth on the moon 
acts as a centripetal or deviating force on the latter, while the 
equal and opposite force acting on the earth may be called 

the centrifugal. If a small block moving on a 
smooth horizontal table is gradually turned from 
its straight course AB by a fixed circular guide, 
tangent to AB at B, the pressure of the guide 
against the block is the centripetal force Mc 2 ~ r 
directed toward the centre of curvature, while 




78 MECHANICS OF ENGINEERING. 

the centrifugal force J/b 2 -^ r is the pressure of the block 
against the guide, directed away from that centre. The cen- 
trifugal force, then, is never found among the forces acting on 
the body whose circular motion we are dealing with. 

The Conical Pendulum, or governor-ball. — Fig. 82. If a 
material point of mass = M = G -f- g, suspended on a cord of 

Jtzu&c length = I, is to maintain a uniform cir- 

\ cular motion in a horizontal plane, with a 

ct\ given radius r, under the action of gravity 

fV and the cord, required the velocity c to be 
^1 B given it. At B we have the body free. 
M """ g y The only forces acting are G and the cord- 

fig. 83. tension P. The sum of their normal com- 

ponents, i.e., 2JV, must = Mc* -— r, i.e., P sin a = Mo 2 -=- r ; 
but, since ^ (vert, comps.) = 0, P cos a = G. Hence 
G tan a = Gc*-^ gr\ .*. c = Vgr tan a. Let u = number of 
revolutions per unit of time, then u = c -£- ^nr = Vg -f- ^ Vh\ 
i.e., is inversely proportional to the (vertical projection)* of 
the cord-length. The time of one revolution is = 1 -f- u. 

Elevation of the outer rail on railroad curves (considera- 
tions of traction disregarded). — Consider a single car as a 
material point, and free, having a given p ^ 

velocity = c. P is the rail-pressure 

against the wheels. So long as the car £ r- — R-^j 

follows the track the resultant R of P j Ny 

and G must point toward the centre of L. ■■ 

curvature and have a value = M& — - y. **\<-- 

But i?= G tan <*, whence tan a = <? 2 -f- g?\ 

If therefore the ties are placed at this 

angle a with the horizontal, the pressure FlG * 84, 

will come upon the tread and not on the flanges of the wheels ; 

in other words, the car will not leave the track. (This is really 

the same problem as the preceding,) 

Apparent weight of a hody at the equator. — This is less than 
the true weight or attraction of the earth, on account of the 
uniform circular motion of the body with the earth in its 
diurnal rotation. If the body hangs from a spring-balance. 




CURVILINEAR MOTION OF A MATERIAL POINT. 79 

whose indication is G lbs. (apparent weight), while the true- 
attraction is G' lbs., we have G f — G = M& -~ r. For M 
we may use G —■ g (apparent values); for r about 20,000,000 
ft.; for c, 25,000 miles in 24 hrs., reduced to feet per second. 
It results from this that G is < G' by -%^-§G' nearly, and 
(since 17 2 = 289) hence if the earth revolved on its axis seven- 
teen times as fast as at present, G would = 0, i.e., bodies- 
would apparently have no weight, the earth's attraction on 
them being just equal to the necessary centripetal or deviating 
force necessary to keep the body in its orbit. 

Centripetal force at any latitude. — If the earth were a ho- 
mogeneous liquid, and at rest, its form would be spherical ; but 
when revolving uniformly about the polar diameter, its form 
of relative equilibrium (i.e., no motion of the particles relatively 
to each other) is nearly ellipsoidal, the polar diameter being an 
axis of symmetry. 

Lines of attraction on bodies at its surface do not intersect 
in a common point, and the centripetal force requisite to keep 
a suspended body in its orbit (a small circle of the ellipsoid), 
at any latitude fi is the resultant, W, of the attraction or true 
weight G' directed (nearly) toward the centre, and of G the 
tension of the string. Fig. 85. G — the apparent weight, in- 
dicated by a spring-balance and MA is its 
line of action (plumb-line) normal to the 
ocean surface. Evidently the apparent 
weight, and consequently g, are less than 
the true values, since i\^must be perpen- , 
dicular to the polar axis, while the true 
values themselves, varying inversely as f^T"^ 

the square of MO, decrease toward the equator, hence the ap- 
parent values decrease still more rapidly as the latitude dimin- 
ishes. The following equation gives the apparent g ior any 
latitude /?, very nearly (units, foot and second): 

g = 32.1808 - 0.0821 cos 2/3. 

(The value 32.2 is accurate enough for practical purposes.' 
Since the earth's axis is really not at rest, but moving aboir 




80 



MECHANICS OF EIN G1JNEEKING. 



the sun, and also about the centre of gravity of the moon and 
earth, the form of the ocean surface is periodically varied, i.e., 
the phenomena of the tides are produced. 

77. Cycloidal Pendulum. — This consists of a material point 
at the extremity of an imponderable, flexible, and inextensible 
cord of length = Z, confined to the arc of a cycloid in a ver- 
tical plane by the cycloidal evolutes shown in Fig. 86. Let 
the oscillation begin (from rest) at A, a height = h above 





\dz 



the vertex. On reaching any lower point, as B (height = z 
above 0), the point has acquired some velocity v, which is at 
this instant increasing at some rate = p t . Now consider the 
point free, Fig. 87 ; the forces acting are P the cord- tension, 
normal to path, and G the weight, at an angle cp with the 
path. From § 74, eq. (5), 2T = Mp t gives 

G cos cp + P cos 90° = (G +■ g)p t ; .\p t = g cos cp 

Hence (eq. (7). § 74). vdv = p t ds gives 

vdv = g cos cpds ; but ds cos cp = — dz ; ,\ vdv = — gdz* 

Summing between A and B, we have 

l— 

the same as if it had fallen freely from rest through the height 
k — 2. {This result evidently applies to any form of path 
when, besides the weight G, there is but one other force, and 
that always normal to the path.) 

From 2JST — Mv 2 ~ n, we have P — G sin cp = Mv 2 -r-r ly 



CURVILINEAR MOTION OF A MATERIAL POINT. 81 

whence P, the cord-tension at any point, may be found (here 
r x = the radius of curvature at any point = length of straight 
portion of the cord). 

To find the time of passing from A to 0, a half-oscillation, 
substitute the above value of v" in v = ds -~ dt, putting ds* 
— dx* + dz*, and we have dtf = (dx 2 + dz 2 ) -+- [2g(h — z)~\. 
To find dx in terms of dz, differentiate the equation of the 
curve, which in this position is 



x = r ver. sin. -1 (z -f- r) -\- V2rz — z* ; 
whence 

rdz (r — z)dz (2r — z)dz 

X ~~ VZrz — z 2 V2rz - z 2 ~~ V2rz — z* y 

.;dx* = [7-1]^' 
(r = radius of the generating circle). Substituting, we have 



*-v1 ( ~ 



Vhz - z* 
lr f^ dz jr F h % . _j z _ Jr 

" V 9^° Vhz—1? ~\ g L vei ' Bm " P~~ V 9' 

Hence the whole oscillation occupies a time = n \/l -+■ g 
(since I = 4r). This is independent of h, i.e., the oscillations 
are isochronal. This might have been proved by showing that 
pt is proportional to OB measured along the curve ; i.e., that 
the motion is harmonic. (§ 59, Prob. 2.) 

78. Simple Circular Pendulum. — If the material point oscil- 
lates in the arc of a circle, Fig. 88, proceeding 
as in the preceding problem, we have finally, j\"*---. ; 
after integration by series, as the time of a full I \ /, 
oscillation, Lr \?-'''4 

1 ■ 1 * jl 9 A 2 . 1 ^ L " x 

"^8 ' I "^256 l 2 "^ J' FlG - 88 - 

Hence for a small h the time is nearly 7t VI -f- g, and the os- 



9 



82 MECHANICS OF ENGINEERING. 

dilations nearly isochronal. (For the Compound Pendulum, 
see § 117.) 

79. Change in the Velocity Square. — From eq. (7), § 74, we 
have -J-(y — c 2 ) — fp t ds. But, from similar triangles, du be- 
ing the projection of any ds of the path upon the resultant 
force R at that instant, Rdu = Tds (or, Prin. of Yirt. Yels. 
§ 62, Rdu = Tds -f JV X 0). T and i^are the tangential and 
normal components of R. Fig. 89. Hence, finally, 

iMv* -\M&-fRdu, (a) 

for all elements of the curve between any two points. In gen- 

^^cis-* t eneral R is different in amount and direc- 

M y--, -"S i — h' — rr> 



^■^'"""'-'i. - v tion for each ds of the path, but du is the 
_\j R ^,. distance through which R acts, in its own 
Fig. 89. direction, while the body describes any ds ; 

Rdu is called the work done by R when ds is described by the 
body. The above equation is read : The difference between the 
initial and final kinetic energy of a body == the work done by 
the resultant force in that portion of the path. 

(These phrases will be further spoken of in Chap. VI.) 
Application of equation (a) to a planet in its orbit about 
the sun. — Fig. 90. Here the only force at any instant is the at- 
traction of the sun R = C -f- u? (see Prob. 3, § 59), 
where C is a constant and u the variable radius 
vector. As u diminishes, v increases, therefore 
dv and du have contrary signs ; hence equation I ^u)\ s 
(a) gives (c being the velocity at some initial ° 
point 0) 

\]£*; -%Mf=-0 M?L = 4----]; (i) 




/ 26T1 1 1 * suw 

.-. v x —\ / c 2 + 17 — .which is independ- fig. 90. 

ent of the direction of the initial velocity c. 

Note. — If u were = infinity, the last member of equation (b) would re- 
duce to G -f- «i, and is numerically the quantity called potential in the 
theory of electricity. 



CURVILINEAR MOTION OF A MATERIAL POINT. 



83 



Application of eg. (a) to a projectile in vacuo 
body's weight, is the only force acting, and 
therefore = E, while M= G ~i- g. There- 
fore equation. (a) gives 



G, the 



Gj dy=G Vl ; 





.-. v, = \& + %gy\<, which is independent of FlG . 91. 

the angle, <*, of projection. 

Application of equation (a) to a body sliding, without fric- 
tion, on a fixed curved guide in a vertical plane ; initial velo- 
city == e at 0. — Since there is some pressure at each point be- 
tween the body and the guide, to consider the body free in 
space, we must consider the guiae removed and that the body 

describes the given curve as a re- 
sult of the action of the two forces, 
its weight G, and the pressure P, 
of the guide against the body. G 
is constant, while P varies from 
point to point, though always (since 
there is no friction) normal to curve. 
At any point, E being the resultant 
of G and P. project ds upon E, thus obtaining du ; on G, 
thus obtaining dy ; on P, thus obtaining zero. But by the 
principle of virtual velocities (see § 62) we have Edu = Gdy 
-j- P X zero = Gdy. which substituted in eq. (a) gives 

c*)=£ , Gay=Gfy y =Gy:, .:v l= V^+2^ 

and therefore depends only on the vertical distance fallen 
through and the initial velocity, i.e., is independent of the 
form of the guide. 

As to the value of P. the mutual pressure between the guide 
and body at any point, since SJV must equal Mv 2 -=- r,r being 
the variable radius of curvature, we have, as in § 77, 

P - G sin <p = Mv 2 h- r ; .. P = £[sin cp-\- (v 2 -f- gr)\ 

As, in general, cp and r are different from point to point of 



G 1 r> 

7^ 




84 MECHANICS OF ENGINEERING. 

the path, P is not constant. (The student will explain how P 
may be negative on parts of the curve, and the meaning of 
this circumstance.) 

80. Projectiles in Vacuo. — A ball is projected into the air 
Y (whose resistance is neglected, hence the 

phrase in vacuo) at an angle = a Q with the 
horizontal ; required its path ; assuming it 
confined to a vertical plane. Resolve the 
motion into independent horizontal (X) 
fig. 93. a nd vertical (Y) motions, G, the weight,, 

the only force acting, being correspondingly replaced by its 
horizontal component = zero, and its vertical component 
= — G. Similarly the initial velocity along X = c x = c cos ar , 
along Y, = c y = csin a a . The X acceleration =p x = -f- M 
= 0, i.e., the X motion is uniform, the velocity v x remains 
= c x = c cos a at all points, hence, reckoning the time from 0, 
at the end of any time t we have 

x = tf(cos a )t (1) 

In the Y motion,^, = (— G) -=- M = — g, i.e., it is uniformly 
retarded, the initial velocity being c y = c sin a ; hence, after, 
any time t, the ^velocity will be (see § 56) v y = c sin a — gt, 
while the distance 

y = c(sin a )t - igf (2) 

Between (1) and (2) we may eliminate t, and obtain as the 
equation of the trajectory or path 

ox* 

y = x tan a. — — - — . 

9 ° 2c cos 2 ex 

For brevity put & = 2^A, h being the ideal height due to the 
velocity c, i.e., & -r- %g (see § 53 ; if the ball were directed ver- 
tically upward, a height h = c* -+- %g would be actually at- 
tained, a being = 90°), and we have 

x* 
y = x tan a, - jj— r^; (3) 

This is easily shown to be the equation of a parabola, with its 
axis vertical. 



CURVILINEAR MOTION OF A MATERIAL POINT. 85 

The horizontal range. — Fig. 94. Putting y = in equa- 
tion (3), we obtain 



x tan a. — —j = — = 0, 

L 4A cos 2 a Q _\ 



which is satisfied both by x = (i.e., at the 

origin), and by x = 4A cos a sin <x . Hence 

the horizontal range for a given c and a is FlG - 94 - 

x r = 4A cos a sin ar = 2A sin 2a . 

For ^ = 45° this is a maximum (c remaining the same), 
being then == 2A. Also, since sin 2a = sin (180° — 2a ) = 
sin 2(90° — a ), therefore any two complementary angles of 
projection give the same horizontal range. 

Greatest height of ascent y that is, the value of y maximum, 
= Vm- — Fig. 94. Differentiate (3), obtaining 

dy _ x 

dx ~~ ° ~~ 2h cos 2 aj 

which, put = 0, gives x = 2h sin a cos a M and this value of 
x in (3) gives y m = h sin 3 a Q . 

(Let the student obtain this more simply by considering the 
T motion separately.) 

To strike a given point ; e being given and a required. — 
Let x' and y' be the co-ordinates of the given point, and a/ 
the unknown angle of projection. Substitute these in equa- 
tion (3), h being known = c 2 -f- 2g, and we have 

x n 1 

y r = x f tan a\ — -.-7 5 7. Put cos 2 a' = - — r— - — = h 

" ° 4A cos 2 a ' ° l _f_ tan 2 a " 

and solve for tan a \ whence 

tan a\ = [2A ± ^4A 2 - x n - 4%'] -i- V. . . (4) 

Evidently, if the quantity under the radical in (4) is negative, 
tan a ' is imaginary, i.e., the given point is out of range with 
the given velocity of projection c= V~2gh; if positive, tan aj 
has two values, i.e., two trajectories may be passed through 
the point ; while if it is zero, tan a Q ' has but one value. 

The envelope, for all possible trajectories having the same 



86 



MECHANICS OF ENGINEERING. 




Fig. 95. 



initial velocity c (and hence the same h) ; i.e., the curve tan- 
gent to them all, has but one point of contact with any one of 
them ; hence each point of the envelope, Fig. 95, must have 

co-ordinates satisfying the con- 
dition, 4A 2 — x n — 4Jiy' = ; i.e. 
(see equation (4) ), that there is 
but one trajectory belonging to it. 
Hence, dropping primes, the 
equation of the envelope is 4A 2 — 
x 2 — 4zhy = 0. Now take 0" as a 
new origin, a new horizontal axis X'\ and reckon y" positive 
downwards ; i.e., substitute x = x" and y = h — y" . The 
equation now becomes a? //2 = AJiy" ; evidently the equation of 
a parabola whose axis is vertical, whose vertex is at 0" 9 and 
whose parameter = 4Ji = double the maximum horizontal 
range. is therefore its focus. 

The range on an inclined plane. — Fig. 96. Let 00 be 
the trace of the inclined plane ; its equation y 
is y = x tan /?, which, combined with the 
equation of the trajectory (eq. 3), will give 
the co-ordinates of their intersection 0. o^f. 
That is, substitute y = x tan ft in (3) and 
solve for x, which will be the abscissa a? 1? of C. This gives 




= tan a — tan /3 == 



sin a sin /? sin (a — /?) 



4A cos a a ~o i cog a ^ cog p cog a ^ cog p y 

.-. x 1 = 4Acos a sin (a — f3) -r- cos /?, and the range 00, 
which = x x -f- cos /3, is = (4A -=- cos 2 /3) cos a sin (ar — /3). (5) 

T 7 ^ maximum range on a given inclined plane, /?, c (and 
.*. A), remaining constant, while a varies. — That is, required 
the value of a which renders 00 a maximum. Differentiat- 
ing (5) with respect to ar , putting this derivative — 0, we have 
[4A -f- cos 2 J3~\ [cos a cos (ar — f3) — sin ar sin (a — /?)] = ; 
whence cos [> + (<x — /?)] = ; i.e., 2a — fi = 90° ; or, 
a = 45° + -J/?, for a maximum range. By substitution this 
maximum becomes known. 

The velocity at any point of the path is v = VvJ -f- *y v a = 



CURVILINEAR MOTION OF A MATERIAL POINT. 



87 



Vc 2 — 2ctg sin a Q -f- g*t* (see the first part of this § 80) ; while 
the time of passage from to any point whose abscissa is x is 
t — x -=- G cos a ; obtained from equation (1). E.g., to reach 
the point B, Fig. 94, we put x = x r = 4A sin a cos or, and ob- 
tain t r = 2c sin <* -=- </. This will give the velocity at 2? =. 

i 7 ? = c. 




Fig. 96a. 



81. Actual Path of Projectiles. — Small jets of water, so long as 
they remain unbroken, give close approximations to parabolic 
paths, as also any small dense object, e.g., a ball of metal, hav- 
ing a moderate initial velocity. The course of a cannon-ball, 
however, with a velocity of 1200 to 1400 feet per second is 
much affected by the resistance of the air, the descending 
branch of the curve being much steeper than the ascending; 
see Fig. 96a. The equation of this curve has not yet been 
determined, but only the expression for the slope (i.e., 
dy : dx) at any point. See Professor Bart- . 
lett's Mechanics, § 151 (in which the body 
is a sphere having no motion of rotation). 
Swift rotation about an axis, as well as an 
unsymmetrical form with reference to the 
direction of motion, alters the trajectory 
still further, and may deviate it from a vertical plane. The 
presence of wind would occasion increased irregularity. See 
Johnson's Encyclopaedia, article " Gunnery." 

82. Special Problem (imaginary ; from Weisbach's Mechan- 
ics. The equations are not homogeneous). — Suppose a ma- 
terial point, mass = 2f, to start 
from the point O, Fig. 97, with 
a velocity = 9 feet per second 
along the — Y axis, being sub- 
jected thereafter to a constant 
attractive X force, of a value X 
= 12M, and to a variable T 
force increasing with the time 
(in seconds, reckoned from O), 




Fig. 97. 



88 MECHANICS OF ENGINEERING. 

viz., Y = SMt. Required the path, etc. For the X motion 
we havej^ — X ~ Jf= 12, and hence 

dv x = I p x dt = 12 / dt ; i.e., v x = 12£; 

/X pt pt 

dx = I v x dt ; i.e., x = 12 / fc& = 6f. . (1) 

For the Y motion^ = Y~ M— St, .-. /' y ^y tf =8 /* fcft ; 

/2/ /it 

dy = I Vydt; 

.-. y = if fdt - %J dt, or y = #» — 9*. . . (2) 

Eliminate t between (1) and (2), and we have, as the equa- 
tion of the path, 

y = ± sW) + %)■- < 3 > 

which indicates a curve of the third order. 
The velocity at any point is (see § 74, eq. (1) ) 



v = Vv; + v; = ±? + 9 (4) 

The length of curve measured from will be (since v — 

ds -r- <#£) 

5 = /*<& =y wft = ^J fdt + 9^ dt = ±f + 92. (5) 

The slope, tan <*, at any point = v y -^ v x = (4£ 2 — 9) -r- 12£, 
<Etan« 4** + 9 

and •'• ~ dT~ = "12?" • (6) 

The radius of curvature at any point (§ 74, eq. (6) ), sub- 
stituting v x — 12t, also from (4) and (6), is 



r = v 



and the normal acceleration — v 2 -^ r (eq. (4), § 74), becomes 
from (4) and (7) p n — 12 (&• P er square second), a constant. 
Hence the centripetal or deviating force at any point, i.e., the 



CURVILINEAR MOTION OF A MATERIAL POINT. 89 

-57V of the forces X and Y, is the same at all points, and = 
Mv* + r = 12M. 

From equation (3) it is evident that the curve is symmetrical 
about the axis X Negative values of t and s would apply to 
points on the dotted portion in Fig. 97, since the body may be 
considered as having started at any point whatever, so long as 
all the variables have their proper values for that point. 

(Let the student determine how the conditions of this motion 
could be approximated to experimentally.) 

83. Relative and Absolute Velocities. — Fig. 98. Let Mbe a 
material point having a uniform motion of velocity v 2 along a 
straight groove cut in the deck of a steamer, which itself has 
a uniform motion of translation, of velocity v x , over the bed of 
a river. In one second M ad- \ / 

vances a distance v 2 along the \ / 

groove, which simultaneously has /^/'" m/W-^*^^--^ 
moved a distance v. — AB with I ■, f//\l 'II ;> ^** > 

the vessel. The absolute path of ^^^^^"'Zlzl^^~-^ — "" 
M during the second is evidently fig. 98. 

w (the diagonal formed on v l and^), which may therefore be 
called the absolute velocity of the body (considering the bed 
of the river as fixed) ; while v 2 is its relative velocity, i.e., rela- 
tive to the vessel. If the motion of the vessel is not one of 
translation, the construction still holds good for an instant of 
time, but v 1 is then the velocity of that point of the deck over 
which M is passing at this instant, and v 2 is Jfef's velocity rela- 
tively to that point alone. 

Conversely, if M be moving over the deck with a given 
absolute velocity — w, v l being that of the vessel, the relative 
velocity v 7 may be found by resolving w into two components, 
one of which shall be v 1 ; the other will be v r 

If w is the absolute velocity and direction of the wind, the 
vane on the mast-head will be parallel to MT, i.e., to v % the 
relative velocity ; while if the vessel be rolling and the mast- 
head therefore describing a sinuous path, the direction of the 
vane varies periodically. 



90 MECHANICS OF ENGINEERING?. 

Evidently the effect of the wind on the sails, if any, will 
depend on v 2 the relative, and not directly on w the absolute, 
velocity. Similarly, if w is the velocity of a jet of water, and 
v 1 that of a water-wheel channel, which the water is to enter 
without sudden deviation, or impact, the channel-partition 
should be made tangent to v 3 and not to w. 

Again, the aberration of light of the stars depends on the 
same construction ; v l is the absolute velocity of a locality of the 
earth's surface (being practically equal to that of the centre) ; 
w is the absolute direction and velocity of the light from a 
certain star. To see the star, a telescope must be directed 
along MT, i.e., parallel to v 2 the relative velocity ; just as in 
the case of the moving vessel, the groove must have the direc- 
tion MT« if the moving material point, having an absolute 
velocity w, is to pass down the groove without touching its 
sides. Since the velocity of light = 192,000 miles per second 
= w 7 and that of the earth in its orbit = 19 miles per second 
= v„ the angle of aberration SMT, Fig. 98, will not exceed 
20 seconds of arc ; while it is zero when w and v 1 are parallel. 

Returning to the wind and sail-boat,' 55 ' it will be seen from 
Fig. 98 that when v 1 = or even > w, it is still possible for v 9 
to be of such an amount and direction as to give, on a sail 
properly placed, a small wind-pressure, having a small fore-and 
aft component, which in the case of an ice-boat may exceed 
the small fore-and-aft resistance of such a craft, and thus v x will 
be still further increased ; i.e., an ice-boat may sometimes travel 
faster than the wind which drives it. This has often been 
proved experimentally on the Hudson River. 

* See § 571 for the mechanics of the sail-boat. 



MOMENT OF INERTIA. 91 



CHAPTER IV. 

MOMENT OF INERTIA. 

[Note. — For the propriety of this term and its use in Mechanics, see 
§§ 114, 216, and 229 ; for the present we deal only with the geometrical 
nature of these two kinds of quantity.] 

85. Plane Figures. — Just as in dealing with the centre of 
gravity of a plane figure (§ 23), we had occasion to sum the 
series fzdF, z being the distance of any element of area, dF, 
from an axis ; so in subsequent chapters it will be necessary to 
know the value of the seriesy^WT^for plane figures of various 
shapes referred to various axes. This summation fz*dF of 
the products arising from multiplying each elementary area of 
the figure by the square of its distance from an axis is called 
the moment of inertia of the plane figure with respect to the 
axis in question • its symbol will be I. If the axis is perpen- 
dicular to the plane of the figure, it may be named the polar 
mom. of inertia (§ 94) ; if the axis lies in the plane, the rec- 
tangular mom. of inertia (§§. 90-93). Since the 7 of a plane 
figure evidently consists of four dimensions of length, it may 
always be resolved into two factors, thus 1= F¥, in which 
F= total area of the figure, while Jc = VI -r- F, is called the; 
radius of gyration, because if all the elements of area were 
situated at the same radial distance, Jc, from the axis, the 
moment of inertia would still be the same, viz., 

I=fk*dF= kfdF= Fh\ 

86. Rigid Bodies. — Similarly, in dealing with the rotary 
motion of a rigid body, we shall need the sum of the series 
fp*dM, meaning the summation of the products arising from 
multiplying' the mass d3f of each elementarv volume dV of a 



92 MECHANICS OF ENGINEERING. 

rigid body by the square of its distance from a specified axis. 
This will be called the moment of inertia of the body with 
respect to the particular axis mentioned (often indicated by a 
subscript), and will be denoted by /. As before, it can often 
be conveniently written Mk 2 , in which M is the whole mass, 
and k its "radius of gyration" for the axis used, k being 
== l//-f- M. If the body is homogeneous, the heaviness, y, of 
all its particles will be the same, and we may write 

I=fp*dM= (y + g)fffdV= (y -r- g) Vh\ 

87. If the body is a homogeneous plate of an infinitely small 
thickness = r, and of area == F, we have T= (y ■—• g)fp 2 dV 
== (y "-f- g)rfp*dF\ i.e., = (y -f- g) X thickness X m#m. *W/*- 
&'# of the plane figure. 

88. Two Parallel Axes. Reduction Formula. — Fig. 99. Let 

Z^ and Z 7 be two parallel axes. Then I z 

=f/fdM, and I z ,=fp ,2 dM. But tf being 

the distance between the axes, so that a 2 

+ & 2 = 6Z 2 , we have p /2 = (x - a)*+(y—by 

= (x 2 + f) + d 2 - 2ax - 2by, and .\ 

I z , =fp*dM+dfd3f- 2afxdM 

-ZbfydM. . (1) 
• — ^ 

fig. 99. But fp*dM = I z ,fdM= M, and from the 

theory of the centre of gravity (see § 23, eq. (1), knowing that 
dM —yd F-f- g, and .'. that (fyd V] -r- g=M) we ImyefxdM 
= Mx andfydM = My ; hence (1) becomes 

l z , = I Z + Mid' - 2ax - %), .... (2) 

in which a and b are the x and y of the axis Z'\ x and y refer 
to the centre of gravity of the body. If Z is a gravity-axis 
(call it <?), both a? and y = 0, and (2) becomes 

A = 7 g + Md* or fe 2 - V +d\ . . (3) 

It is therefore evident that the mom. of inertia about a grav- 
ity-axis is smaller than about any other parallel axis. 

Eq. (3) includes the particular case of a plane figure, by 




MOMENT OF INERTIA. 



93 



writing area instead of mass, i.e., when Z (now g) is a gravity- 



axis. 



T z ,=I g -\-Fd\ 



(4) 



89. Other Reduction Formulae ; for Plane Figures. — (The axes 
here mentioned lie in the plane of the figure.) For two sets 
of rectangular axes, having the same origin, the following holds 
good. Fig. 100. Since 

I x =ftfdF, and I T =fx*dF, 

we have I x + I T -f(a? + y*)dF. 

Similarly, I n + I v =f{tf + u')dF. 

But since the x and y of any dFh&Ye the same hypothenuse as 
the u and v, we have v 2 -f- u 2 = a? 2 -|- y 2 ; . \ I x -f- I Y — Ijj -f- I r . 





Fig. 100. 



Fig. 100a. 



Let Xbe an axis of symmetry ; then, given I x and I 7 {O is 
anywhere on X). required I v , TJ being an axis through O and 
making any angle a with X. 

Itt =ffdF=f(y cos a — x sin otfdF\ i.e., 

Ijj == cos 2 (xfifdF— 2 sin a cos afxydF-\- sin 2 afx 2 dF. 

But since the area is symmetrical about X, in summing up the 
products xydF, for every term x( -f- y)dF, there is also a term 
a?( — y)dF to cancel it ; which gives fxydF = 0. Hence 

Ijj = cos 2 al x -|- sin 2 ^7" F . 

The student may easily prove that if two distances a and b 
be set off from O on X and Y respectively, made inversely 
proportional to Vl x and Vl T , and an ellipse described on a and 
b as semi-axes ; then the moments of inertia of the figure about 



94 



MECHANICS OF ENGINEERING. 



any axes through are inversely proportional to the squares 
of the corresponding semi-diameters of this ellipse ; called 
therefore the Ellipse of Inertia. It follows therefore that the 
moments of inertia about all gravity-axes of a circle, or a 
regular polygon, are equal ; since their ellipse of inertia must 
be a circle. Even if the plane figure is not symmetrical, an 
"ellipse of inertia" can be located at any point, and has the 
properties already mentioned ; its axes are called the principal 
axes for that point. 

90. The Rectangle. — First, about its base. Fig. 101. Since 
all points of a strip parallel to the base 

— .^..^ ■ * fe-> have the same co-ordinate, z, we may take 

dz the area of such a strip for dF = bdz; 

■ g Ah, ph 

.'.I B =J z*dF=bJ z'dz 



•f 

! 


» 


A j 

M 

dz j 
B 






Z 

* 


t 


— i 



Fig. 101. 



Fig. 102. 



» 



M> 



/Secondly, about a gravity-axis parallel to base. 

dF= bdz .-. I g =fz*dF= bj ' z 2 dz = -^bh\ 

Thirdly, about any other axis in its plane. Use the results 
already obtained in connection with the reduction-formulae of 

§§88, 89. 

90a. The Triangle. — First, about an axis through the vertex 
and parallel to the base ; i.e., I v 
in Fig. 103. Here the length 
of the strip is variable ; call it y. ^ 
From similar triangles 
y = (b -=- h)z\ 



— b 



h~ 




Fig. 103. 



Fig. 104. 



/. I v =fz'dF= fz*ydz = {b~ h)f z z dz = \bh\ 

Secondly, about g, a gravity -axis parallel to the base. Fig, 
104. From § 88, eq. (4), we have, since F— ibh and 

d = %h, I g = I v - Fd> = ibh 5 - ibh . %W = ^bh\ 



MOMENT OF INERTIA. 



95 



Thirdly, Fig. 104, about the base ; I B =\ From § 88, eq. 
(4), I B = I g -\- Fd% with d == ih ; hence 

h = sW + V& • V* 2 = tW- 

91. The Circle. — About any diameter, as g, Fig. 105. Polar 
co-ordinares, I g = fz*dF. Here we take dF= area of an ele- 
mentary rectangle = pdcp . dp, while z = p sin cp. 

* & » 







L 



ffci 



B 



Fig. 105. 



Fig. 106. 



I g = J I (p sin cpfpdcpdp = / sin 2 cpdcpj p s dp\ 

= — / sin 2 ^<£> = — / 4(1 •— cos 2cp)dcp 

r* p 2 *\~\ 1 "I 

= -j^ [jj <*<P - 4 • cos 2<M2<p)J 

1 r 27r /l 1 \ ' 

o)-(0-0)]. i.e.,I g = l*r\ 



2tt 



92. Compound Plane Figures. — Since I — fz^dF is an in- 
finite series, it may be considered as made up of separate 
groups or subordinate series, combined by algebraic addition, 
corresponding to the subdivision of the compound figure into 
component figures, each subordinate series being the moment of 
inertia of one of these component figures ; but these separate 
moments must all be referred to the same axis. It is con- 
venient to remember that the (rectangular) I of a plane 
figure remains unchanged if we conceive some or all of its 
elements shifted any distance parallel to the axis of refer- 
ence. E.g., in Fig. 106, the sum of the I B of the rectangle OF, 
and that of FD is = to the I B of the imaginary rectangle 



96 



MECHANICS OF ENGINEERING. 



formed by shifting one of them parallel to B, until it touches 
the other ; i.e., I B of OF+ I B of FD = ^bji, 3 (§ 90). Hence 
the I B of the T shape in Fig. 106 will be = I B of rectangle 
AD - I B of rect. OF - I B of rect. FD. 

That is, I B of T = ii?A' - &A 8 ]- • • • (§ 90). . . (1) 

About the gravity-axis, g, Fig. 106. To find the distance d 
from the base to the centre of gravity, we may make use of 
eq. (3) of § 23, writing areas instead of volumes, or, experi- 
mentally, having cut the given shape out of sheet-metal or 
card-board, we may balance it on a knife-edge. Supposing d 
to be known by some such method, we have, from eq. (4) of 
§ 88, since the area F= bh — bji^ I g = I B — Fd 2 ; 

i.e., I g = i[M 3 - Z>A 3 ] ~ (M ~ \K)d\ . . (2) 
The double-T (orx), and the box forms of Fig. 106&, if 
--* *>- ► symmetrical about the gravity- 

axis g, have moments of inertia 
alike in form. Here the grav- 
ity-axis (parallel to base) of the 
compound figure is also a grav- 
ity axis (parallel to base) of each 
of the two component rectangles, of dimensions b and A, h x and 
A x , respectively. 

Hence by algebraic addition we have (§ 90), for either com- 
pound figure, 

l g =UW-lAl (3) 

(If there is no axis of symmetry parallel to the base we must 
proceed as in dealing with the T-form.) Similarly for the ring,. 















If, 


i h 




-t&r 






•f-Hbj— 


&. | 






— »■■' 




h 


—L- 















Fig. 106a. 





Fig. 107. Fig. 108. 

Fig. 107, or space between two concentric circumferences, we 
have, about any diameter or g (§ 91), 

/ 9 = i*('', 4 -n 4 ) (*> 



MOMENT OF INERTIA. 97 

The rhombus about a gravity-axis, g, perpendicular to a 
diagonal, Fig. 108. — This axis divides the figure into two 
equal triangles, symmetrically placed, hence the I g of the 
rhombus equals double the moment of inertia of one triangle 
about its base ; hence (§ 90#) 

I Q = 2 . ^(W = &W (5) 

(The result is the same, if either vertex, or both, be shifted 
any distance parallel to AB.) 

For practice, the student may derive results for the trapezoid ; 
for the forms in Fig. 106, when the inner corners are rounded 
into equal quadrants of circles; for the double-T*, when the 
lower flanges are shorter than the upper; for the regular 
polygons, etc. 

93. If the plane figure be bounded, wholly or partially, by 
curves, it may be subdivided into an infinite number of strips, 
and the moments of inertia of these (referred to the desired 
axis) added by integration, if the equations of the curves are 
known ; if not, Simpson's Rule, for a finite even number of 
strips, of equal width, may be employed for an approximate 
result. If these strips are parallel to the axis, the 1 of any one 
strip = its length X its width X square of distance from axis; 
while if perpendicular to, and terminating in, the axis, its 
/ = \ its width X cube of its length (see § 90). 

A graphic method of determining the moment of inertia of 
any irregular figure will be given in a subsequent chapter. 

94. Polar Moment of Inertia of Plane Figures (§ 85). — Since 
the axis is now perpendicular to the plane of the figure, inter- 
secting it in a point, 0, the distances of the ele- ; 

merits of area will all radiate from this point, 

and would better be denoted by p instead of z ; 

hence, Fig. 109, /p 2 ^ 7 is the polar moment of 

inertia of any plane figure about a specified 

point ; this may be denoted by I p . But p 2 Fig. 109. 

= a? 2 -f- ?/ 2 , for each dF\ hence 

/, =/<y + y*)dF=fx*dF+ftfdF = I r + I z . 




98 



MECHANICS OF ENGINEERING. 



i.e., the polar moment of inertia about any given point in 
the plane equals the sum of the rectangular moments of iner- 
tia about any two axes of the plane figure, which intersect at 
right angles in the given point. We have therefore for the 
circle about its centre 

I p — iTtr* + Jtzt 4 = ^nr i ; 
For a ring of radii r x and r„ 

h = i« - <) ; 

For the rectangle about its centre, 

I* = T W + tV« 3 = iJ>W + #) ; 

For the sqiiare, this reduces to 

A V Q U • 

(See §§90 and 91.) 

95. Slender, Prismatic, Homogeneous Rod. — Returning to the 
moment of inertia of rigid bodies, or solids, we begin with that 

of a material line, as it might be called, about 
an axis through its extremity making some an- 
gle a with the rod. Let I = length of the rod, 
Fits cross-section (very small, the result being 
strictly true only when F = 0). Subdivide 
the rod into an infinite number of small prisms, 
each having F as a base, and an altitude = ds. Let y = the 
heaviness of the material ; then the mass of an elementary 
prism, or dlf, = (y -=- g)Fds, while its distance from the axis 
Z is p = s sin a. Hence the moment of inertia of the rod 
with respect to Z as an axis is 

I z = J t p 2 dM= (y -r- g)F sin* afs 2 ds = i(y + g)Fl z sin 2 or. 

But yFl -T- g = mass of rod and I sin a = a, the distance of 
the further extremity from the axis ; hence I z = \Ma? and 
the radius of gyration, or Jc, is found by writing -J Ma' 2 = MW ; 
.-. F = \a\ or h = Via (see § 86). If a = 90°, a = I. 

96. Thin Plates. Axis in the Plate. — Let the plates be homo- 
geneous and of small constant thickness == r. If the surface of 




MOMENT OF INERTIA. 99 

the plate be = F, and its heaviness y, then its mass = yFr - a 
From § 87 we have for the plate, about any axis, ' 

J- (r 1 9)rX mom. of inertia of the plane figure formed by 

the shape of the plate J. 

i Rectangular plate. Gravity-axis parallel Vto.-Dimen 
sions 5 and h. From eq. (1) and § 90 we have 

Similarly, if the base is the axis, I B = 1J/A 2 • & 2 — £#» 
Triangular plate. Axis through vertex parallel to base- 
-brom eq. (1) and § 90a, dimensions being b and h, 
f v = (y + g)rlW = (yibhr ~- g)^ = %Mtf- .-. ^ = ^ 

Circular plate, with any diameter as axis.— From ea m 
and fc 91 we have 4 ' v ; 

^ = \Y + g)r\nr* = {ynr'r ~- g)fr* = i Mr *. ^ ¥ = ^ 

97. Plates or Kight Prisms of any Thickness (or Altitude) 
Axis Perpendicular to Surface (or Base).-As before, the solid is 
homogeneous, i.e., of constant heaviness y 
let the altitude = h. Consider an elementary ^+\ 

pnsm, Fig. Ill, whose lengtli is parallel to the s£Z4**) 
axis of reference Z. Its altitude = h = that ^H^-f-^ 
of the whole solid; its base = dF = an element 4 \\\ p J^ \\ 

of * the area of the base of solid ; and each ^^^H ^ 

point of it has the same p. Hence we may n* 111 
take its mass, = yhdF ~ g, as the dM in summing the series 
2 z =fp>dM; 

-h={yh~g)f P \iF 

= (M -*- ?) X polar mom. of inertia of base. . . (2) 

folWin e . USe ° f ^ ^ "^ thG reSUltS iD §94 We 0btain the 

o * 

Circular plate, or ryft «W^ cylinder, about the R eo- 
metrical axis. ri= radius, A = altitude. 

Might parallelopiped or rectangular plate.— -Kg. 112 



luo 



MECHANICS OF ENGINEERING. 



For a hollow cylinder, about its geometric axis, 
I g = (yh+gftnir; - O = \M{^ + rft .: ¥ = J(r,' + r^ 





Fig. 112. Fig. 113. 

98. Circular Wire. — Fig. 113 (perspective). Let Z be a 
gravity-axis perpendicular to the plane of the wire ; X and Y 
lie in this plane, intersecting at right angles in the centre 0. 
The wire is homogeneous and of constant (small) cross-section. 
Since, referred to Z, each dM has the same p — r, we have 
l z —f r 2 dM= Mr\ Now I x must equal I Y , and (§ 94) their 
sum = I z , 

.*. I x , or I T , = %Mr% and k x % or Jc Y * = %r\ 

99. Homogeneous Solid Cylinder, about a diameter of its base, 
— Fig. 114. I x = ? Divide the cylinder into an infinite num- 
ber of laminse, or thin plates, parallel to the 
base. Each is some distance z from X, of 
thickness dz, and of radius r (constant). In 
each draw a gravity-axis (of its own) parallel to 

fig. 114. X. We may now obtain the I x of the whole 

cylinder by adding the / x 's of all the laminse. The I g of any 
one lamina (§96, circular plate) = its mass X i^ 2 ; hence its 
I x (eq. (3), § 88) = its I g -f- (its mass) X £ a . Hence for the 
whole cylinder 

Ix = J* \kydznr' - g\V + *•)] 

= Wy -5- g) [i^T dz +f* z*dz~] ; 

i.e, I x = {nfhy ~ g){^ + W) = *{& + W- 

100. Let the student prove (1) that if Fig. 114 represent 
any right prism, and k F denote the radius of gyration of any 
one lamina, referred to its gravity-axis parallel to X, then the 
I x of whole prism = M{Jc F -\- -J-A 2 ); and (2) that the moment 




MOMENT OF INERTIA. 



101 




of inertia of the cylinder about a gravity-axis parallel to the 
base is = M^r' + -^tf). 

101. Homogeneous Right Cone. — Fig. 115. First, about an 
axis V, through the vertex and parallel to the base. As before, 
divide into laminae parallel to the base. Each is a 
circular thin plate, but its radius, x, is not = r, but, f 
from proportion, is x = (r -r- h)z. 

The I oi any lamina referred to its own gravity- f 
axis parallel to J 7 " is (§96) = (its mass) X i# 2 , and £ W L 
its / F (eq. (3), §88) is .*. = its mass X ia? 2 + Fig. ns. 
its mass X s 2 . 

Hence for the whole cone, 

I v = /* (jztfdzy -s- ff)fof + s°] 

V 

Secondly, about a gravity-axis parallel to the base. — From 
eq. (3), §88, with d = |A (see Prob. 7, §26), and the result 
just obtained, we have /= M^\r' 2 -\- JA 2 ]. 

Thirdly, about its geometric axis, Z. — Fig. 116. Since the 
axis is perpendicular to each circular lamina through the centre, 
its I z (§ 97) is 

= its mass X i(rad.) 2 — (yntfdz -& g)^- 
Now x = (r -J- h)z, and hence for the whole cone 




Fig. 116. 





102. Homogeneous Right Pyramid of Rectangular Base. — 

About its geometrical axis. Proceeding as in the last para- 



102 



MECHANICS OF ENGINEERING. 



graph, we derive I z = Mfod?, in which d is the diagonal of the 
base. 



103. Homogeneous Sphere. — About any diameter. Fig. 118. 
I z = ? Divide into laminae perpendicular to Z. By § 97, and 
noting that x 2 = r 2 — z\ we have finally, for the whole sphere, 



Iz = {y* -*- 2g) 



+ r 



(*+* -&* + &) = -fry*** *g 




Fig. 119 



For a segment, of one or two bases, put proper limits for z 
in the foregoing, instead of + r and — r. 

104. Other Cases. — Parabolic plate, Fig. 119, homogeneous 
Y and of (any) constant thickness, about 

^v an axis through 0, the middle of the 
-X chord, and perpendicular to the plate. 
This is 

Fio. 120. /^Jffl^ + fA 2 ). 

The area of the segment is — %hs. 

For an elliptic plate, Fig. 120, homogeneous and of any 
constant thickness, semi-axes a and b, we have about an axis 
through O, normal to surface I = M^\a? -f- & 2 ] ; while for a 
very small constant thickness 

I x =Mib% and I Y =Mia\ 

The area of the ellipse = nab. 

Considering Figs. 119 and 120 as plane figures, let the 
student determine their polar and rectangular moments of 
inertia about various axes. 

(For still other cases, see p. 518 of Rankine's Applied 
Mechanics, and pp. 593 and 594 of Coxe's Weisbach.) 

105. Numerical Substitution. — The moments of inertia of 
plane figures involve dimensions of length alone, and will be 
utilized in the problems involving flexure and torsion of beams, 
where the inch is the most convenient linear unit. E.g., the 



MOMENT OF INERTIA. 



103 



polar moment of inertia of a circle of two inches radius about 
its centre is \7ir* = 25.13 -f- biquadratic, or four-dimension, 
inches, as it may be called. Since this quantity contains four 
dimensions of. length, the use of the foot instead of the inch 
would diminish its numerical value in the ratio of the fourth 
power of twelve to unity. 

The moment of inertia of a rigid body, or solid, however, 
3= M Jc* = (G — $)%?, in which G, the weight, is expressed in 
units of force, g involves both time and space (length), while ¥ 
involves length (two dimensions). Hence in any homogeneous 
formula in which the I of a solid occurs, we must be careful to 
employ units consistently ; e.g., if in substituting G -=- g for M 
(as will always be done numerically) we put g = 32.2, we 
should use the second as unit of time, and the foot as linear 
unit. 

106. Example. — Kequired the moment of inertia, about the 
axis of rotation, of a pulley consisting of a rim, four parallelo- 
pipedical arms, and a cylindrical hub which may be considered 
solid, being filled by a portion of the shaft. 
Fig. 121. Call the weight of the.hub G, 
its radius r ; similarly, for the rim, G 2 , r x 
and r 9 ; the weight of one arm being = G t . 
The total /will be the sum of the /'s of 
the component parts, referred to the same 
axis, viz. : Those of the hub and rim will 
be (G + g)ir> and (G 2 ~ fifr? + <), 
respectively (§ 97), while if the arms are FlG - 121 « 

not very thick compared with their length, we have for them 
(§§95 and 88) 

± (0H- 0) [K'. - r)* - l(r, - rf + [r + ftr, - r)Y], 
as an approximation (obtained by reduction from the axis at 
the extremity of an arm to a parallel gravity-axis, then to the 
required axis, then multiplying by four). In most fly-wheels, 
the rim is proportionally so heavy, besides being the farthest 
removed from the axis of rotation, that the moment of inertia 
of the other parts may be for practical purposes neglected. 




104 



MECHANICS OF ENGINEERING. 



107. Ellipsoid of Inertia. — The moments of inertia about 
all axes passing through any given point of any rigid body 
whatever may be proved to be inversely proportional to the 
squares of the diameters which they intercept in an imaginary 
ellipsoid, whose centre is the given point, and whose position 
in the body depends on the distribution of its mass and the 
location of the given point. The three axes which contain the 
three principal diameters of the ellipsoid are called the Princi- 
pal Axes of the body for the given point. This is called the 
ellipsoid of inertia. (Compare § 89.) Hence the moments of 
inertia of any homogeneous regular polyedron about all gravity- 
axes are equal, since then the ellipsoid becomes a sphere. It 
can also be proved that for any rigid body, if the co-ordinate 
axes JT, JT, and Z, are taken coincident with the three principal 
axes at any point, we shall have 

fxydM = ; fyzdM = ; and fsxdM = 0. 



DYNAMICS OF A RIGID BODY. 105 



CHAPTER V. 
DYNAMICS OF A RIGID BODY 

108. General Method. — Among the possible motions of a 
rigid body the most important for practical purposes (and for- 
tunately the most simple to treat) are : a motion of translation, 
in which the particles move in parallel right lines with equal 
accelerations and velocities at any given, instant ; and rotation 
about a fixed axis, in which the particles describe circles in 
parallel planes .with velocities and accelerations proportional 
(at any given instant) to their distances from the axis. Other 
motions will be mentioned later. To determine relations, or 
equations, between the elements of the motion, the mass and 
form of the body, and the forces acting (which do not neces- 
sarily form an unbalanced system), the most direct method to 
be employed is that of two equivalent systems of forces (§ 15), 
one consisting of the actual forces acting on the body, con- 
sidered free, the other imaginary, consisting of the infinite 
number of forces which, applied to the separate material points 
composing the body, would account for their individual mo- 
tions, as if they were an assemblage of particles without mutual 
actions or coherence. If the body were at rest, then considered 
free, and the forces referred to three co-ordinate axes, they 
would constitute a balanced system, for which the six summa- 
tions 2X, 2Y, 2Z, ^(mom.) x , 2(mom.) Y , and I(mom.) z . 
would each = ; but in most cases of motion some or all of 
these sums are equal (at any given instant), not to zero, but to 
the corresponding summation of the imaginary equivalent 
system, i.e., to expressions involving the masses of the particles 
(or material points), their distribution in the body, and the 



106 



MECHANICS OF ENGINEERING. 



elements of the motion. That is, we obtain six equations by 
putting the ^X of the actual system equal to the 2X of the 
imaginary, and so on ; for a definite instant of time (since some 
of the quantities may be variable). 



109. Translation. — Fig. 122. At a given instant all the par- 
ticles have the same velocity = v, in parallel right lines (par- 
allel to the axis X, say), and the 
same acceleration p. Required 
Mp the 2X of the acting forces, 
dM shown at (I.). (II.) shows the 
imaginary equivalent system, con- 
sisting of a force = mass X ace. 
= dMp applied parallel to X to 
each particle, since such a force 
would be necessary (from eq. (TV.) 
§ 55) to account for the accelerated rectilinear motion of the 
particle, independently of the others. Putting (2X )j= (2J) n , 




(X) (ID 

Fig. 122. 



we have 



(2X) I =fpdM=pfdM = Mp. . 



(V.) 



It is evident that the resultant of system (II.) must be paral- 
lel to X\ hence that of (I.), which = (^X)j and may be de- 
noted by i?, must also be parallel to X\ let a = perpendicular 
distance from It to the plane YX\ a will be parallel to Z. 
Now put [2(mom.) r ]j = [2(mom. y)]ii, (l^is an axis perpen- 
dicular to paper through 0) and we have — JRa = —fdMpz 
= —pfdMz = — pMz (§88), i.e., a = z. A similar result 
may be proved as regards y. Hence, if a rigid body has a 
motion of translation, the resultant force must act in a line 
through the centre of gravity (here more properly called the 
centre of mass), and parallel to the direction of motion. Or, 
practically, in dealing with a rigid body having a motion of 
translation, we may consider it concentrated at its centre of 
mass. If the velocity of translation is uniform, R = M X 
= 0, i.e., the forces are balanced. 



DYNAMICS OF A RIGID BODY. 107 

110. Rotation about a Fixed Axis. — First, as to the elements 
of space and time involved. Fig. 123. Let O be the axis of 
rotation (perpendicular to paper), OY & fixed d -«-—^ >. w 
line of reference, and OA a convenient line of <^~~ ^~/~\\ 
the rotating body, passing through the axis and / / a ) 
perpendicular to it, accompanying the body in I — — J~ 
its angular motion, which is the same as that of V^^ — - — ' 
OA. Just as in linear motion we dealt with FlG - 123 - 
linear space («§), linear velocity (V), and linear acceleration (p),. 
so here we distinguish at any instant ; 
a, the angular space between Y and OA ; 

gLoc 
go = the angular velocity, or rate at which a is changing ; 

and 

dco oVa . 

u = -J- = ~^, the angular acceleration, or rate at which go 

is changing. 

These are all reckoned in ^--measure and may be + or — , 
according to their direction against or with the hands of a 
watch. 

(Let the student interpret the following cases : (1) at a cer- 
tain instant go is -|-, and 6 — ; (2) go is — , and 6 -)-; (3) a is 
— , go and 6 both +; (4) a -\-. go and 6 both — .) For rotary 
motion we have therefore, in general, 



da ,^ TV _ doo d 2 <x 



go = 



(VI.) S„=— ;. . (VII.) 



dt 9 v ' dt dt 



and .-. Godco =6da; (YIII.) 

corresponding to eqs. (I.), (II.), and (III.) in § 50, for rectilinear 
motion. 

Hence, for uniform rotary motion, go being constant and 
6 = 0, we have a = Got, t being reckoned from the instant 
when a = 0. 

For uniformly accelerated rotary motion 6 is constant, and 



108 



MECHANICS OF ENGINEERING. 



if go denote the initial angular velocity (when a and t =s 0), 

we may derive, precisely as in §56, 

go = go, + 6t ; . . (1) a = otf + ietf; . . (2) 

g = a?, 2g <a ° , > * ' (3 ) and « = 4K + ^ • • W 

If in any problem in rotary motion 6*, go, and a have been 
determined for any instant, the corresponding linear values for 
.any point of the body whose radial distance from the axis is p, 
will be s = ap (= distance described by the point measured 
along its circular path from its initial position), v = Gop = its 
velocity, andj^ ,== Op its tangential acceleration, at the instant 
in question. 

Examples. — (1) What value of go, the angular velocity, is 
implied in the statement that a pulley is revolving at the rate 
of 100 revolutions per minute ? 

100 revolutions per minute is at the rate of 2?r X 100 
= 628.32 (^-measure units) of angular space per minute 
= 10.472 per second ; .\ go — 628.32 per minute or 10.472 
per second. 

(2) A grindstone whose initial speed of rotation is 90 revo- 
lutions per minute is brought to rest in 30 seconds, the an- 
gular retardation (or negative angular acceleration) being con- 
stant ; required the angular acceleration, 0, and the angular 
space a described. Use the second as unit of time. 
g? u = 27r-jj-# = 9.4248 per second ; .*. from eq. (1) 

GO — Gtf 

e = — — ^ 



= — 9.424 -=- 30 = — 0.3141 (^-measure units) 

per " square second." The angular space, from eq. (2) is 

a = aof + idf = 30 X 9.42 - £(0.314)900 =: 141.3 
(7r-measure units), i.e., the stone has made 22.4 revolutions in 
coming to rest and a point 2 ft. from the axis has described a 
distance s= ap = 141.3 X 2 = 282.6 ft. in its circular path. 

111. Rotation. Preliminary Problem. Axis Fixed. — For 
clearness in subsequent matter we now consider the following 



DYNAMICS OP A RIGID BODY. 



109 




Fig. 124. 



simple case. Fig. 124 shows a rigid body, consisting of a 
drum, an axle, a projecting arm, all 
of which are imponderable, and a 
single material' point, whose weight 
is G and mass M. An imponderable 
flexible cord, in which the tension is 
kept constant and = P, unwinds 
from the drum. The axle coincides 
with the vertical axis Z, while the cord 
is always parallel to Y. Initially (i.e., when t = 0) M lies at 
rest in the plane ZY. Required its position at the end of any 
time t (i.e., at any instant) and also the reactions of the bearings 
at and O x , supposing no vertical pressure to exist at O x , and 
that P and M are at the same level. JSTo friction. At any in- 
stant the eight unknowns, a, go, 6, X , Y , Z , JT T , and Y x , may 
be found from the six equations formed by putting 2X, etc., 
of the system of forces in Fig. 124, equal, respectively, to the 
2X, etc., of the imaginary equivalent system in Fig. 125, and 
two others to be mentioned subsequently. Since, at this in- 
stant, the velocity of M must be v = Gop and its tangential ac- 
celeration p t = 6p, its circular motion 
could be produced, considering it free (eq. 
(5), § 74), by a tangential force T = mass 
X Pt — MOp, and a normal centripetal 
force N=Mv* -f- p=M{gdpY + p=Go*Mp. 
Hence the system in Fig. 125 is equivalent 



b 
Fig. 125. 

to that of Fig. 124, and from putting the ^ (mom.) z of one 
= that of the other, we derive 



Pa = Tp ; i.e., Pa — 0Mp\ 



(1) 



whence 6 becomes known, and is evidently constant, since P, 
a, M, and p are such. .*. the angular motion is uniformly ac- 
celerated, and from eqs. (1) and (2), § 110, go and a become 
known ; 

i.e., Go=0t 9 . . . (2) and a = %dt\ .... (3) 
Putting [2Z of 124) = (2Z of 125), gives 

Z. - G= 0; i.e., Z n = G .... (4) 



110 MECHANICS OF ENGINEERING. 

Proceeding similarly with the ^2X of each system, 
X -\-X 1 = Tcos a — iTsin a = 6Mp cos a — co 2 Mp&m or, (5) 
and with the 2 Y of each, 
P + Y + T x = - Tsintf-T^costf^ - OMpsina 

— c*? 2 jffp cos a ; (6) 
while with the 2 (mom.) x we have, conceiving all the forces in 
each system projected on the plane ZT"(see §38), and noting 
that y = p cos a and x = p sin a, 

+ Gp cos a + Y x l + P6 = - (OMp sin a)h—(G?Mp cosa)b,(7) 
and with the 2 (mom.) r , 
— Gp sin a — X X Z = — (6Mp cos a)5 + {co 2 Mp sin a)5. . (8) 

From (7) we may find Y x \ from (8), X x \ then X and I 7 ",, 
from (5) and (6). It will be noted that as the motion proceeds 
6 remains constant ; go increases with the time, ol with the 
square of the time ; Z Q is constant, = G ; while X^ Y , X x , 
and 5^ have variable values dependent on p cos a and p sin <*, 
i.e., on the co-ordinates y and x of the moving material point. 

112. Particular Supposition in the Preceding Problem with 
Numerical Substitution. — Suppose we have given (using the 
foot-pound-second sy stein of units in which g = 32.2) G = 64 .4 
lbs., whence 
M=(G + g) = 2; P = 4 lbs., 1 = 4 ft., ft = 2 ft., a = 2 ft., 
and p = 4 ft.; and that J/ is just passing through the plane 
ZX, i.e., that « = \n. We obtain, first, the angular accelera- 
tion, eq. (1), 

6 = Pa + Mp 2 = 8 ^ S2= 0.25 = J. 
From eqs. (2) and (3) we have at the instant mentioned (not- 
ing that when a was = 0, t was = 0) 

go* = 2a6 = %tt = 0.7854 -f , 
while (2) gives, for the time of describing the quadrant, 
t = go -r- = 3.544. . . . seconds. 
Since at this instant cos a = and sin a ■— 1, we have, from 

(7), 

+ 0H-F,X4 + 4X2^ -JX2X4X2; .-. I> - 3 lbs. 



DYNAMICS OF A KIGID BODY. 



Ill 



The minus sign shows it should point in a direction contrary 
to that in which it is drawn in Fig. 124. Eq. (8) gives 

- 64.4 x 4-Xx X 4=-0+}?rx2x4x2 \.\X X = - 67.54 lbs. 
And similarly, knowing Y x and X x , we have from (5) and (6), 
X = + 61.26 lbs., and Y = - 3.00 lbs. 
The resultant of X x and Y v also that of X , Y , and Z , can 
now be found by the parallelogram (and parallelopipedon) of 
forces, both in amount and position, noting carefully the direc- 
tions of the components. These resultants are the actions of 
the supports upon the ends of the axle ; their equals and 
opposites would be the actions or pressures of the axle against 
the supports, at the instant considered (when M is passing 
through the plane ZX\ i.e., with a = \7t). (At the same in- 
stant, suppose the string to break; what would be the effect on 
the eight quantities mentioned?) 

113. Centre of Percussion of a Rod suspended from one End. — 

Fig. 126. The rod is initially at rest (see (I.) in figure), is straight, 
homogeneous, and of constant 



*«<L- y - 



o 



"e 



(i.) 



dT 



(ID 
Fig. 126. 



(in.) 
(II.) shows the rod 



(small) cross-section. Neglect its 
weight. A horizontal force or 
pressure, P, due to a blow (and 
varying in amount during the 
blow), now acts upon it from the 
left, perpendicularly to the axis, 
Z, of suspension. An accelerated 
rotary motion begins about the fixed axis Z. 
free, at a certain instant, with the reactions X and Y put in 
at O . (III.) shows an imaginary system which would produce 
the same effect at this instant, and consisting of a dT = dMdp, 
and a dJV = ctfdMp applied to each dM, the rod being composed 
of an infinite number of dM\ each at some distance p from 
the axis. Considering that the rotation has just begun, go, the 
angular velocity is as yet small, and will be neglected. Ee- 
quired Y the horizontal reaction of the support at in term* 
of P. By putting 2. Y n = 2 Y UI , we have 

P -Y, =fdT= 6fpdM= 6M~p. 



112 



MECHANICS OF ENGINEEKING. 



.*. Y = P — OM p ; p is the distance of the centre of gravity 
from the axis (N.B. J'pdM = M ' p is only true when all the 
p's are parallel to each other). Bat the value of the angular 
acceleration 6 at this instant depends on P and a, for 2 (mom.)^ 
in (II.) = 2 (mom.)* in (III.), whence Pa = Of/fdM = dI Zl 
where I z is the moment of inertia of the rod about Z, and from 
§ 95 = \Ml\ Now p = il ; hence, finally, 

L 2 * ij 

If now Y v is to = 0. i.e., if there is to be no shock between 
the rod and axis, we need only apply P at a point whose dis- 
tance a-=.\l from the axis ; for then Y = 0. This point is 
called the centre of percussion for the given rod and axis. It 
and the point of suspension are interchangeable (see § 118). 
(Lay a pencil on a table; tap it at a point distant one third of 
the length from one end ; it will begin to rotate about a vertical 
axis through the farther end. Tap it at one end ; it will begin 
to rotate about a vertical axis through the point first mentioned. 
Such an axis of rotation is called an axis of instantaneous rota- 
tion, and is different for each point of impact — just as the 
point of contact of a wheel and rail is the one point of the 
wheel which is momentarily at rest, and about which, therefore, 
all the others are turning for the instant. Tap the pencil at 
its centre of gravity, and a motion of translation begins ; see 
§ 109.) 



114. Rotation. Axis Fixed. 




General Formulae, 
z 

/Tx 



•Consider 




Fig. 127. Fig. 128. 

ing now a rigid body of any shape whatever, let Fig. 127 indi- 
cate the system of forces acting at any given instant, Z being 



I 



DYNAMICS OF A KIGID BODY. 113 

the fixed axis of rotation, go and 6 the angular velocity and 
angular acceleration, at the given instant. X and Y are two 
axes, at right angles to each other and to Z, fixed in space. At 
this instant each dM of the body has a definite x, y, and cp 
(see Fig. 128), which will change, and also a p, and z, which will 
not change, as the motion progresses, and is pursuing a circu- 
lar path with a velocity = cop and a tangential acceleration 
= Op. Hence, if to each dM of the body (see Fig. 128) we 
imagine a tangential force dT — dM&p and a normal force 

— dM{oopf -^ p — czfdMp to be applied (eq. (5), § 74), and 
these alone, we have a system comprising an infinite number of 
forces, all parallel to XY, and equivalent to the actual system 
in Fig. 127. Let ^X, etc., represent the sums (six) for Fig. 
127, whatever they may be in any particular case, while for 
128 we shall write the corresponding sums in detail. Noting 
that 

fdN cos cp = cofdMp cos cp — cofdMy = ^My,{% 88); 
that fdN sin cp = oofdMp sin q> — oofdMx = gd*Mx; 
and similarly, that fdT cos cp = OfdMp cos cp = 6My, and 
fdT sin cp = 6Mx; while in the moment sums (the moment 
of dT cos cp about Y, for example, being — dT cos cp . z — 

— SdMp (cos cp)z= — 6dMyz, the sum of the moms. Y of all the 
(dT cos 0>)'s = - SfdMyz) 

fdT 'cos cpz — Of dMyz, fdX sin cpz — aofdMxz, etc., 
we have, since the systems are equivalent, 

2X = + 6 My - co 2 Mx; .... (IX.) 
2Y = - 6Mx~ co*My; . . . . (X.) 

2Z= 0; (XL) 

J2 moms.* = - 6/dMxz - cofdMyz ; . (XII.) 

2 moms. 7 = - dfdMyz + oofdMxz ; . (XIII.) 

2 moms. z = OfdMp' = 6I Z . . • . (XIY.) 

These hold good for any instant. As the motion proceeds x 

and y change, as also the sums fdMxz and fdMyz. If the 

body, however, is homogeneous, and symmetrical about the 

'plane XY, fdMxz and fdMyz would always = zero ; since 

9 



114 



MECHANICS OF ENGINEERING. 



the z of any dM does not change, and for every term dMy(-\-z\ 
there would be a term dMy(— z) to cancel it; similarly for 
fdMxz. The eq. (XIV.), I(moms. about axis of rotat.) — 
fdTp = OfdMp 2 = {angular accel.) X (mom. of inertia of 
body about axis of rotat. \ shows how the sum fdMp 2 arises in 
problems of this chapter. That a force dT — dMdp should 
be necessary to account for the acceleration (tangential) dp of 
the mass dM, is due to the so-called inertia of the mass (§ 54), 
and its moment dTp, or OdMp 2 , might, with some reason, be 
called the moment of 'inertia of the dM, ^ndfddMp 2 — OfdMp 2 
that of the whole body. But custom has restricted the name 
to the sum fdMp 2 , which, being without the 6, has no term to 
suggest the idea of inertia. For want of a better the name is 
still retained, however, and is generally denoted by /. (See 
§§86, etc.) ' / 



115. Example of the Preceding. — A homogeneous right par- 
allelopiped is mounted on a vertical 
axle (no friction), as in figure. is 
at its centre of gravity, hence both 
x and y are zero. Let its heaviness 
be y, its dimensions h, b xl and b (see 
§ 97). XY is a plane of symmetry, 
hence both fdMxz and fdMyz are 
zero at all times (see above). The 
tension P in the (in extensible) cord 
is caused by the hanging weight P x 
(but is not = P x , unless the rotation is uniform). The figure 
shows both rigid bodies free. P y will have a motion of trans- 
lation ; the parallelopiped, one of rotation about a fixed axis. 
No masses are considered except P x -\- g, and bhb x y -=- g. The 
I z = Mkz of the latter = its mass X T V(^, 3 + &), § 97. At 
any instant, the cord being taut, if jp — linear acceleration of 
P 15 we have 

da eq. (a) 




Fig. 129. 



P 

From (XIV.), Pa = 6I Z ; .\ P = 6I Z 



(1) 



DYNAMICS OF A RIGID BODY. 115 

For the free mass P x -^ g we have (§109) P x — P = 
mass X ace., 

= (P, + g)p = (P, - y)ft* ; .-, P = P,(l - ©a + ?). (2) 

Equate these two values of P and solve for 0, whence 

All the terms here are constant, hence & is constant ; there- 
fore the rotary motion is uniformly accelerated, as also the 
translation of P v The formulae of § 56, and (1), (2), (3), and 
(4) of § 110, are applicable. The tension P is also constant ; 
see eq. (1). As lor the five unknown reactions (components) 
at O x and 0„ the bearings, we shall find that they too are con- 
stant ; for 

from (IX.) we have X x + X % = ; (4) 

from (X.) we have P + Z, + T, = ; (5) 

from (XI.) we have Z 2 - G = ; (6) 

from (XII.) we have P . AO+ Y l .7\O-Y t .UJ) = 0;(T) 
from (XIII.) we have - X x . Oft + X 3 . Oft = 0. (8) 

Numerical substitution in the above problem. — Let the par- 
allelopiped be of wrought-iron ; let P x == 48 lbs.; a = 6 in. = 
i ft.; 5 = 3 in. = I ft, ' (see Fig. JL12) ; b ± =_2 ft. 3 in. =t £ ft.; 
and h = 4 in. = J ft. Also iet Oft = 0,0 = 18 in. = f ft., 
and AO = 3 in. = ^ ft. Selecting the foot-pound-second 
system of units, in which g = 32.2, the linear dimensions must 
be used in feet, the heaviness, y, of the iron must be used in 
lbs. per cubic foot, i.e., y = 480 (see § 7), and all forces in lbs., 
times in seconds. 

The weight of the iron will be G == Vy = bbjiy = -J- . f- . -J 
X 480 = 90 lbs.; its mass = 90 -f- 32.2 = 2.79 ; and its mo- 
ment of inertia about Z — I z — Mk z * = M^(b* + b*) — 2.79 
X 0.426 = 1.191. (That is, the radius of gyration, Tc z , = 
V0A26 = 0.653 ft.; or the moment of inertia, or any result 
depending solely upon it, is just the same as if the mass were 
concentrated in a thin shell, or a line, or a point, at a distance 
of 0.653 feet from the axis.) We can now compute the an- 
gular acceleration, 6, from eq. (3) ; 



116 



MECHANICS OF ENGINEERING. 



e = 



48 X* 



= 15.36 



1.191 + (48 -^ 32.2) X i ~ 1.191 + 0.372 
7r-measure units per " square second." The linear acceleration 
of P x \&p — 6a = 7.68 feet per square second for the uniform, 
ly accelerated translation. 

Nothing has yet been said of the velocities and initial condi- 
tions of the motions ; for what we have derived so far applies 
to any point of time. Suppose, then, that the angular velocity 
go = zero when the time, t = ; and correspondingly the ve- 
locity, v = ooa, of translation of P„ be also = when t = 0. 
At the end of any time t, go = 6t (§§ 56 and 110) and v =pt 
= dat ; also the angular space, a = \Qtf, described by the par- 
allel opiped during the time t, and the linear space s = ipf 
= idatf, through which the weight P 1 has sunk vertically. 
For example, during the first second the parallelopiped has ro- 
tated through an angle a = \Bf = \ X 15.36x1 = 7.68 units, 
^-measure, i.e., (7.68 -=- 2?r) = 1.22 revolutions, while P 1 has 
sunk through s = \Qat = 3.84 ft., vertically. 

The tension in the cord, from (2), is 

P = 48(1 - 15.36 X i -5- g) = 48(1 - 0.24) = 36.48 lbs. 

The pressures at the bearings will be as follows, at any in- 
Btant : from (4) and (8), X l and X 2 must individually be zero ; 
from (6) Z 2 = <?= Vy = 90 lbs.; while from (5) and (7), T x 
t= — 21.28 lbs., and Y % — — 15.20 lbs., and should point in a 
direction opposite to that in which they were assumed in Fig. 
129 (see last lines of § 39). 

116. Torsion Balance. A Variably Accel. Rotary Motion. 

Axis Fixed. — A homogeneous solid having an axis of symmetry 

is suspended by an elastic prism, 
or filament (whose mass may be 
neglected), so that the latter is 
vertical and coincident with the 
axis of symmetry, and is not only 
supported, but prevented from 
turning at its upper extremity. 
fig. 130. If the solid is turned about its 

axis away from its position of rest and set free, the torsional 




DYNAMICS OF A RIGID BODY. 117 

elasticity of the rod or filament, which is fixed in the solid, 
causes an oscillatory rotary motion. Required the duration of 
an oscillation. Fig. 130. 

Take the axis Y at the middle of the oscillation (the original 
position of rest). Reckon the time from the instant of passing 
this position. Let the initial angular velocity = oo . As the 
motion progresses go diminishes, i.e., 6 is negative. 

To consider the body free, conceive the rod cut close to the 
body (in which it is firmly inserted), and in the section thus 
exposed put in the vertical tension P 1 , and also the horizontal 
forces forming a couple to which at any instant the twisting 
action (of the portion of rod removed upon the part left in the 
free body) is known to be due. Call the moment of this couple 
Qb (known as the moment of torsion) ; it is variable, being 
directly proportional to the angle a ; hence, if by experiment 
it is found to be = QJ) V when a is = a x , for any value of a it 
will be Qb = (Q 1 b 1 -~ a^a = Ga, in which G is the constant 
factor. 

At any instant, therefore, the forces acting are G, P' , and 
those equivalent to the couple whose moment = Qb = Ga. 
(No lateral support is required ; the student would find the X x , 
Y x , X 2 , and Y 2 of Fig. 129 to be individually zero, if put in ; 
remembering that here, x and y both = 0, as also folMxz and 
fdMyz ; and that the forces of the couple will not be repre- 
sented in any of the six summations of § 114, except in 
^E moms. z ) 

From eq. (XIV.), § 114, we have - Qb, i.e., - Ga, = 6I Z , 
from which 

6 — — (<7-t- I z )a, or, for short, 6 — — Ba. . . (1) 

Since B is constant, and there is an initial (angular) velocity 
=s g? , and since the variables 6, go, and a, in angular motion 
correspond precisely to those (j>,v, and s) of rectilinear motion, 
it is evident that the present is a case of harmonic motion, 
already discussed in Problem 2 of § 59. Applying the results 
there obtained, since B of eq. (1) corresponds to the a of that 
problem, we find that the oscillations are isochronal, i.e., their 



118 



MECHANICS OF ENGINEERING. 



durations are the same whatever the amplitude (provided the 
elasticity of the rod is not impaired), and that the duration of 
one oscillation (from one extreme position to the other) is 
f = it 4- VB] or finally, 



t' = n \/aJ z -Z Qfi v 



(2) 



117. The Compound Pendulum is any rigid body allowed to 
oscillate without friction under the action of gravity when 
mounted on a horizontal axis. Fig. 131 shows the 
body free, in any position during the progress of 
the oscillation. C is the centre of gravity; let OC 
= s. From (XIY.), § 114, we have 2 (mom. about 
fixed axis) 

= angul. ace. X mom. of inertia. 




.'. — Gs sin a — 6I Q , 
and 6 = — Gs sin a -^- L = 



Mgs sin a -f- MJc\, 



i.e., 



e 



gs sin oc 



(i) 



Hence d is variable, proportional to sin a. Let us see what 
the length I = OJT, of a simple circular pendulum, must be, to. 
have at this instant (i.e., for this value of a) the same angular 
acceleration as the rigid body. The linear (tangential) accelera- 
tions of K, the extremity of the required simple pendulum 
would be (§ 77) jp t — — 9 s ^ n a * an d hence its angular accelera- 
tion would = — g sin a -i- 1. Writing this equal to 6 in eq. 
(1), we obtain 



l = K 



(2) 



But this is independent of ol ; therefore the length of the sim- 
ple pendulum having an angular acceleration equal to that of 
the oscillating body is the same in all positions of the latter, 
and if the two begin to oscillate simultaneously from a position 
of rest at any given angle a x with the vertical, the}' will keep 
abreast of each other during the whole motion, and hence have 



DYNAMICS OF A RIGID BODY. 119 

the same duration of oscillation ; which is .*. , for small ampli- 
tudes (§ 78), 



f ±=z 7t VT^-g = 7t Vk 2 -T- gs, . . . . (3) 

K is called the centre of oscillation corresponding to the given 
centre of suspension 0, and is identical with the centre of per- 
cussion (§113). 

Exam/pie. — Required the time of oscillation of a cast-iron 
cylinder, whose diameter is 2 in. and length 10 in., if the axis 
of suspension is taken 4 in. above its centre. If we use 32.2 
for <7, all linear dimensions should be in feet and times in 
seconds. From § 100, we have 

Ic = M(ir* + ^) = M(i.^ + ^.m) = M J h-W' 
From eq. (3), § 88, 

I = I c + Ms' = M[^ . W + i] = Mx 0.170; 



.-. Jc * = 0.170 sq. ft.; .\ t'— n 4/0.170 -r- (32.2 Xi) == 0.395 sec. 

118. The Centres of Oscillation and Suspension are Inter- 
changeable. — (Strictly speaking, these centres are points in the 
line through the centre of gravity perpendicular to the axis of 
suspension.) Refer the centre of oscillation K to the centre 
of gravity, thus (Fig. 132, at (I.) ) : 



Mk 2 __ Mk c * + Ms 2 _ _kc 
Ms s ~ Ms ~ s ~ s 



ail n, ail n c -p atlh rv c , . 

Sl = l- S = - 1 r r - - S = m 8 = — . (1) 



Now invert the body and suspend it at K\ , . , , 

required CK^ or s 2 , to find the centre of \~"".j9\ \ K *~ \\s x 
oscillation corresponding to K as centre of | s / \ \ j\ 
suspension. By analogy from (1) we have sj] * \ \„ /jj^ 
s 2 = kc -=- s ± ; but from (1), k c 2 -5- s 1 = s .*. / — — — i \——l 
s 2 = s ; in other words, K x is identical with (i.) (ii.) 

O. Hence the proposition is proved. FlG - m 

Advantage may be taken of this to determine the length L 
of the theoretical simple pendulum vibrating seconds, and thus 
finally the acceleration of gravity from formula (3), § 117, viz., 



120 MECHANICS OF ENGINEERING. 

when if = 1.0 and I (now = L) has been determined experi- 
mentally, we have 

g (in ft. per sq. second) s= L (in ft.) X tt 2 . . . (2) 

This most accurate method of determining g at any locality 
requires the use of a bar of metal, furnished with a sliding 
weight for shifting the centre of gravity, and with two project- 
ing blocks provided with knife-edges. These blocks can also 
be shifted and clamped. By suspending the bar by one knife- 
edge on a proper support, the duration of an oscillation is com- 
puted by counting the total number in as long a period of 
time as possible; it is then reversed and suspended on the 
other with like observations. By shifting the blocks between 
successive experiments, the duration of the oscillation in one 
position is made the same as in the other, i.e., the distance be- 
tween the knife-edges is the length, I, of the simple pendulum 
vibrating in the computed time (if the knife-edges are not equi- 
distant from the centre of gravity), and is carefully measured. 
The I and t' of eq. (3), § 117, being thus known, g may be com- 
puted. Professor Bartlett gives as the length of the simple 
pendulum vibrating seconds at any latitude /? 

Z(in feet) = 3.26058 - 0.008318 cos 2/3. 

119. Isochronal Axes of Suspension. — In any compound 
pendulum, for any axis of suspension, there are always three 
others, parallel to it in the same gravity plane, for which the 
oscillations are made in the same time as for the first. For 
any assigned time of oscillation t 1 , eq. (3), § 117, compute the 
corresponding distance CO = s of O from O; 

i.e., from t = n -^-- = y - , 



we have s = (gt' i +27t 2 )± VigT^n*) - h c \ . . (1) 

Hence for a given t' , there are two positions for the axis O 
parallel to any axis through O, in any gravity-plane, on both 
sides; i.e., four parallel axes of suspension, in any gravity- 
plane, giving equal times of vibration ; for two of these axes 



DYNAMICS OF A RIGID BODY. 



121 



we must reverse the body. E.g., if a slender, homogeneous, 
prismatic rod be marked off into thirds, the (small) vibrations 
will be of the same duration, if the centre of suspension is 
taken at either extremity, or at either point of division. 

Example. — Required the positions of the axes of suspension, 
parallel to the base, of a right cone of brass, whose altitude is 
six inches, radius of base, 1.20 inches, and weight per cubic inch 
is 0.304 lbs., so that the time of oscillation may be a half- 
second. (K.B. For variety, use the inch-pound-second system 
of units, first consulting § 51.) 

120. The Fly- Wheel in Fig. 133 at any instant experiences 
a pressure P' against its crank-pin from the connecting-rod 
and a resisting pressure P" from the teeth of a spur-wheel with 




wmzmmw- 



in' 



Fig 133. 



which it gears. Its weight G acts through G (nearly), and 
there are pressures at the bearings, but these latter and G have 
no moments about the axis C (perpendicular to paper). The 
figure shows it free, P" being assumed constant (in practice 
this depends on the resistances met by the machines which D 
drives, and the fluctuation of velocity of their moving parts). 
P\ and therefore T its tangential component, are variable, 
depending on the effective steam-pressure on the piston at any 
instant, on the obliquity of the connecting-rod, and in high- 
speed engines on the masses and motions of the piston and con- 
necting-rod. Let r =3 radius of crank-pin circle, and a the 
perpendicular from {/on P" . From eq. (XI Y.), § 114, we 
have 

Tr - P"a = 6I C , /. 6=(Tr- P"a) -=- I c , . (1 



122 MECHANICS OF ENGINEERING. 

as the angular acceleration at any instant ; substituting which in 
the general equation (Till.), § 110, we obtain 

I c oodGo = Trda - P"ada (2) 

From (1) it is evident that if at any position of the crank-pin 
the variable Tr is equal to the constant P"a, 6 is zero, and 
consequently the angular velocity go is either a maximum or a 
minimum. Suppose this is known to be the case both at m 
and n ; i.e., suppose T, which was zero at the dead-point A y 
has been gradually increasing, till at n, Tr = P"a ; and there- 
after increases still further, then begins to diminish, until at m 
Tr again = P"a, and continues to diminish toward the dead- 
point B. The angular velocity go, whatever it ma}^ have been 
on passing the dead-point A, diminishes, since 6 is negative, 
from A to n, where it is gd w a minimum ; increases from n to 
m, where it reaches a maximum value, ao m . n and m being 
known points, and supposing oo n known, let us inquire what 
co m will be. From eq. (2) we have 

I c J n GDdoj=J n Trda-P"j n ada. . . (3) 

But rda = ds = an element of the path of the crank-pin, and 
also the " virtual velocity" of the force T, and ada = ds", an 
element of the path of a point in the pitch-circle of the fly- 
wheel, the small space through which P" is overcome in dt. 
Hence (3) becomes 

^ciOm 2 - <»n) =J n Tds - P" X linear arc EF. (4> 

To determine / Tds we might, by a knowledge of the vary- 
ing steam-pressure, the varying obliquity of the connecting-rod, 
etc., determine T for a number of points equally spaced along 
the curve nm, and obtain an approximate value of this sum by 
Simpson's Rule ; but a simpler method is possible by noting 
(see eq. (1), § 65) that each term Tds of this sum = the corre- 

^ponding term Pdx in the series / Pdx, in which P = the 



DYNAMICS OF A KIGID BODY. 123 

effective steam-pressure on the piston in the cylinder at any in- 
stant, dx the small distance described by the piston while the 
crank-pin describes any ds, and n' and m' the positions of the 
piston (or of cross-head, as in Fig. 133) when the crank-pin is 
at n and m respectively. (4) may now be written 

Icii^m - c*n) =J Pdx - P" X linear arc FF, (5) 

from which Go m may be found as proposed. More generally, it 
is available, alone (or with other equations), to determine any 
one (or more, according to the number of equations) unknown 
quantity. This problem, in rotary motion, is analogous to that 
in § 59 (Prob. 4) for rectilinear motion. Friction and the in- 
ertia of piston and connecting-rod have been neglected. As 
to the time of describing the arc nm, from equations similar to 
(5), we may determine values of go for points along nm, divid- 
ing it into an even number of equal parts, calling them g^, go„ 
etc., and then employ Simpson's Rule for an approximate value 

C m p m da 
t— I — (from eq. (VL), § 110) ; e.g., with 

four parts, we would have 



r m l 

t = — (angle nOm, 7r-meas.) 
Ln 1^ 



.GO n GO ' GO ' GO ' GO m _A 



.(6) 



121. Numerical Example. Fly-Wheel.— (See Fig. 133 and 
the equations of § 120.) Suppose the engine is non-condensing 
and non-expansive (i.e., that P is constant), and that 

P = 5500 lbs., r = 6in. = ift., a = aft., 

and also that the wheel is to make 120 revolutions per minute, 
i.e., that its mean angular velocity is to be 

a/ = i^j. x 27r, i.e., go' = ^n " radians" per sec. 

First, required the amount of the resistance P" (constant) 
that there shall be no permanent change of speed, i.e., that the 
angular velocity shall have the same value at the end of a com- 
plete revolution as at the beginning. Since an equation of the 
form of eq. (5) holds good for any range of the motion, let 



124 



MECHANICS OF ENGINEERING. 



that range be a complete revolution, and we shall have zero as 
the left-hand member ; /Pdx = P X 2 f t. = 5500 lbs. X 2 ft., 
or 11,000 foot-pounds (as it may be called); while P" is un- 
known, and instead of lin. arc FFwe have a whole circumfer- 
ence of 2 ft. radius, i.e., 4tt ft.; 

,\ = 11,000 - P" X 4 X 3.1416; whence P" = 875 lbs. 
Secondly, required the proper mass to be given to the fly- 
wheel of 2 ft. radius that in the forward stroke (i.e., while the 
crank-pin is describing its upper semicircle) the max. angular 
velocity ao m shall exceed the minimum oo n by only 1 L q(*>\ assum- 
ing (which is nearly true) that i(oo m -f- co n ) = go'. There be- 
ing now three unknowns, we require three equations, which 
are, including eq. (5) of § 120, viz.: 

Mh c ^{GD m + G0 n ){0J m - OD n ) 



Jn' 



Pdx - P" X linear arc EF\ (5) 

i(GD m + GD n )= GO'= 4:7! ; (7) and OD m - GD n = ^Gd' = *7t. (8) 

The points n and m are found most easily and with sufficient 
accuracy by a graphic process. Laying off the dimensions to 
scale, by trial such positions of the crank- pin are found that 
T, the tangential component of the thrust P' produced in the 
connecting-rod by the steam-pressure P (which may be resolved 
into two components, along the connecting-rod and a normal 
to itself) is =(a -f- r)P", i.e., is == 3500 lbs. These points will 
be n and m (and two others on the lower semicircle). The 
positions of the piston n' and m', corresponding to n and m of 
the crank-pin, are also found graphically in an obvious manner. 
We thus determine the angle nCm to be 100°, so that linear 
arc FF = \%%7t X 2 ft. = ifrn 



ft., while 



/ Pdx = 

t/n' 



I 



5500 lbs. X / dx= 5500x^W=5500x 0.77ft., 



n'm! being scaled from the draft. 



Now substitute from (7) and (8) in (5), and we have, with 
k c = 2 ft. (which assumes that the mass of the fly-wheel is con- 
centrated in the rim), 



DYNAMICS OF A KIGID BODY. 125 

(G ± g) X 4 X 4:7t x \n = 5500 X 0.77 - 875 X *#*', 
which being solved for G (with g = 32.2 ; since we have used 
the foot and second), gives G = 600.7 lbs. 

The points of max. and min. angular velocity on the back- 
stroke may be found similarly, and their values for the fly- 
wheel as now determined ; they will differ but slightly from 
the co m and ao n of the forward stroke. Professor Cotterill says 
that the rim of a fly-wheel should never have a max. velocity 
> 80 ft. per sec; and that if made in segments, not more than 
40 to 50 feet per second. In the present example we have for 
the forward stroke, from eqs. (7) and (8), Gd m = 13.2 (7T-measure 
units) per second; i.e., the corresponding velocity of the wheel- 
rim is v m = oo m a = 26.4 feet per second. 

122. Angular Velocity Constant. Fixed Axis. — If go is con- 
stant, the angular acceleration, 6, must be = zero at all times,, 
which requires 2 (mom.) about the axis of y ^ 

rotation to be = (eq. (XIV.), § 114). An X V^C" 
instance of this occurs when the only forces <^J ss^^. 
acting are the reactions at the bearings on Nl A 

the axis, and the body's weight, parallel to g /* f G 
or intersecting the axis ; the values of these /KJ/^ 
reactions are now to be determined for dif- / * — S 
ferent forms of bodies, in various positions fig. 134. 

relatively to the axis. (The opposites and equals of these reac- 
tions, i.e., the forces with which the axis acts upon the bearings, 
are sometimes stated to be due to the " centrifugal forces" or 
" centrifugal action," of the revolving body.) 

Take the axis of rotation for Z, then, with 6 = 0, the equa- 
tions of § 114 reduce to 

2X= -co'Mx; .... (IXa.) 
2Y = —Go*My; .... (Xa.) 

2Z = ; (XIa.) 

2 moms. x = — oofdMyz ; . . . (Xlla.) 
2 raoms.y = -f- oofdMxz ; . . . (XUIa.) 
2 moms.2 = (XlVa.) 



126 



MECHANICS OF ENGINEEKING. 



-P- 




For greater convenience, let us suppose the axes JTand Y 
(since their position is arbitrary so long as they are perpen- 
dicular to each other and to Z) to revolve with the body in its 
uniform rotation. 

122a. If a homogeneous body have a plane of symmetry 
and rotoite uniformly about any axis Z perpendicular to that 
plane {intersecting it at 0), then the acting forces are equiva- 
lent to a single force, = otfMp, applied 
at and acting in a gravity-line, but 
directed away from the centre of 
gravity. It is evident that such a 
force P = co^Mp, applied as stated 
FlG - 185 (see Fig. 135), will satisfy all six con- 

ditions expressed in the foregoing equations, taking X through 
the centre of gravity, so that x = p. For, from (IX&.), P must 
= GD 2 Mp, while in each of the other summations the left- 
hand member will be zero, since P lies in the axis of X\ and 
as their right-hand members will also be zero for the present 
body (y = ; and each of the sums fdMyz and fdlfxz is zero, 
since for each term dMy{ -\- z) there is another dMy{ — z) 
to cancel it ; and similarly, for fdMxz), they also are satisfied ; 
Q.E.D.* Hence a single point of support at will suffice to 
maintain the uniform motion of the body, and the pressure 
against it will be equal and opposite to P. 

First Example. — Fig. 136. Supposing (for greater safety) 
that the uniform rotation of 210 revolutions 
per minute of each segment of a fly- wheel is 
maintained solely by the tension in the cor- 
responding arm, P ; required the value of P 
if the segment and arm together weigh -^ of 
a ton, and the distance of their centre of 
gravity from the axis is p = 20 in., i.e., = f ft. 
ton-second system of units, with g 

P = co*Mp = [^ X 2tt] 2 X DhV -*" 32 - 2 ] X i = °- 83 tons < 
or 1660 lbs. 



Fig. 136. 



With the foot- 
32.2, we have 



That is, neglecting gravity. 



DYNAMICS OF A RIGID BODY. 



12? 




Second Example. — Fig. 137. Suppose the uniform rotation 
of the same fly-wheel depends solely on the tension in the rim, 
required its amount. The figure shows the half- <- f 
rim free, with the two equal tensions, P\ put in at 
the surfaces exposed. Here it is assumed that the 1 -_C 
arms exert no tension on the rim. From §122$ we 
have 2P' = oo 2 Mp, where Mis the mass of the half- p' 
rim, and p its gravity co-ordinate, which may be ob- fig. 137. 
tained approximately by § 26, Problem 1, considering the rim 
as a circular wire, viz., p = 2r -f- tt. 

Let M — (180 lbs.) -f- g, with r = 2 ft. We have then 

P' = i(22) 2 (180 +. 32.2)(4 -mt) = 1718.0 lbs. 

(In reality neither the arms nor the rim sustain the tensions 
just computed ; in treating the arms we have supposed no duty 
done by the rim, and vice versa. The actual stresses are less, 
and depend on the yielding of the parts. Then, too, we have 
supposed the wheel to take no part in the transmission of mo- 
tion by belting or gearing, which would cause a bending of the 
arms, and have neglected its weight.) 

122b. If a homogeneous body have a line of symmetry and 
rotate uniformly about an axis parallel to it (0 being the foot 
of the perpendicular from the centre of gravity on the axis), 
then the acting forces are equivalent to a single force P 
= od'Mp. applied at and acting in a gravity-line away 

from the centre of gravity. 

Taking the axis X. through the 

du centre of gravity, Z being the 

H axis of rotation, Fig. 138, while 

Z' is the line of symmetry, pass 
an auxiliary plane Z ' Y' parallel 
to ZT. Then the sum fdMxz 
may be written fdM(p + x')z 
which = ~pfdMz + fdMx'z. 
But fdMz = Mz = 0, since ~z 
= 0, and every term dM(-\~ x')z is cancelled by a numerically 




■X~~ 



Fig. 138. 



128 



MECHANICS OF ENGINEERING 



equal term dM{— x')z of opposite sign. Hence fdMxz = 0. 
AhofdMyz = 0, since each positive product is annulled by an 
equal negative one (from symmetry about Z'). Since, also r 
y = 0, all six conditions in § 122 are satisfied. Q. E. D. 

If the homogeneous body is any solid of revolution whose 
geometrical axis is parallel to the axis of rotation, the forego- 
ing is directly applicable. 

122c. If a homogeneous body revolve uniformly about any 
axis lying in a plane of symmetry, the acting forces are equiv- 
alent to a single force P — oo 2 Mp, acting parallel to the grav- 
ity-line which is perpendicular to the axis (Z), and away 
from the centre of gravity, its distance from any origin in 
the axis Z being — [fdMxz] -^ Mp {the plane ZX being a 
gra/vity -plane). — Fig. 139. From the position of the body we 
have p = x, and y = ; hence if a 
value Gj*Mp be given to P and it be 
made to act through Z and parallel to* 
X, and away from the centre of gravity, 
all the conditions of § 122 are satisfied 
except (XII#.) and (Xllla.). But 
symmetry about the plane XZ makes 
fdMyz = 0, and satisfies (Klla.), and 
by placing P at a distance a —fdMxz -^ Mp from along Z 
we satisfy (Xllla.). Q. E. D. 

Example. — A slender, homogeneous, prismatic rod, of length 
= I, is to have a uniform motion, about a ver- q, 
tical axis passing through one extremity, 
maintained by a cord-connection with a fixed 
point in this axis. Fig. 140. Given go, cp, I, 
(p — \l cos cp), and F the cross-section of the 
rod, let s = the distance from to any dM 
of the rod, dM being = Fyds -^ g. The x 
of any dM = s cos cp ; its z = s . sin cp ; 

/.fdMxz = (Fy -r- g) sin cp cos cp J s*ds 

*J 

— \(Fyl ~r- g)V sin cp cos cp = \Ml* sin cp cos cp. 




Fig. 139. 




Fig. 140. 



DYNAMICS OF A RIGID BODY. 129 

Hence a, =fdMxz -f- Mp, is = \l sin <p, and the line of ac- 
tion of P ( = crfMp = go 2 (Fyl -f- ^) J£ cos 9?) is therefore 
higher up than the middle of the rod. Find the intersection 
D of G and the horizontal drawn through Z at distance a from 
0. Determine P' by completing the parallelogram GP' , at- 
taching the cord so as to make it coincide with P'\ for this w T ill 
satisfy the condition of maintaining the motion, when once be- 
gun, viz., that the acting forces G, and the cord-tension P\ 
shall be equivalent to a force P = co^Mp, applied horizontally 
through Z at a distance a from 0. 

123. Free Axes. Uniform Rotation. — Referring again to § 122 
and Fig. 134, let us inquire under what circumstances the 
lateral forces, X x , Y x , X < Y , with which the bearings press 
the axis, to maintain the motion, are individually zero, i.e., that 
the hearings are not needed, and may therefore he removed 
(except a smooth horizontal plane to sustain the body's weight), 
leaving the motion undisturbed like that of a top " asleep." 
For this, not only must *2X and 2 Y both be zero, but also 
(since otherwise X 1 and X might form a couple, or Y 1 and Y 
similarly) 2 (moms.) z and 2 (moms.) r must each = zero. The 
necessary peculiar distribution of the body's mass about the 
axis of rotation, then, must be as follows (see the equations of 
§122): 

First, x and y each = 0, i.e., the axis must he a gravity -axis. 

Secondly, fdMyz = 0, and fdMxz = 0, the origin being any- 
where on Z, the axis of rotation. 

An axis (Z) (of a body) fulfilling these conditions is called 
a Free Axis, and since, if either one of the three Principal Axes 
for the centre of gravity (see § 107) be made an axis of rotation 
(the other two being taken for X and Y), the conditions 
x = 0, ~y — 0, fdMxz = 0, and fdMyz = 0, are all satisfied, 
it follows that every rigid hody has at least three free axes, 
which are the Principal Axes of Inertia of the centre of 
gravity at right angles to each other. 

In the case of homogeneous hodies free axes can often be 
determined by inspection : e.g., any diameter of a sphere ; any 



130 



MECHANICS OF ENGINEERING. 



transverse diameter of a right circular cylinder through its 

centre of gravity, as well as its geometrical axis ; the geomet- 
rical axis of any solid of revolution ; etc. 

124. Rotation about an Axis which has a Motion of Translation. 

— Take only the particular case where the moving axis is a 
<jt\^ gravity-axis. At any instant, let the 

dU dP velocity and acceleration of the axis be v 
and p ; the angular velocity and accelera- 
te tion about that axis, go and 6. Then, since 




Fig. 141. 



*"P the actual motion of a dMin any dt is 
compounded of its motion of rotation 
about the gravity-axis and the motion of 
translation in common with that axis, 
we may, in forming the imaginary equiva- 
lent system in Fig. 141, consider each dM as subjected to the 
simultaneous action of dP = dMp parallel to X, of the tan- 
gential dT = dMOp, and of the normal dJV '= dM(aop)* -i- p 
= cxfdMp. Take Xm the direction of translation, Z (perpen- 
dicular to paper through 0) is the moving gravity-axis ; Y 
perpendicular to both. At any instant we shall have, then, the 
following conditions for the acting forces (remembering that 
p sin cp = y.fdMy = My = ; etc.) : 

2X =fdP -fdT sin cp -fdXcos <p = Mp; . (1) 

2Y=fdTcos cp -fdXmi cp =0;. . (2) 

2 moms., =fdTp -fdPy = OfdMp 2 = 6I Z = 6Mk z \ (3) 

and three other equations not needed in the following example. 
Example. — A homogeneous solid of revolution rolls {with- 
out slipping) down a rough inclined 
plane. Investigate the motion. Con- 
sidering the \>oo\y free, the acting forces 
are G (known) and JV and P, the un- x 
known normal and tangential compo- *P 
nents of the action of the plane on the 
roller. If slipping occurs, then P is the 
sliding friction due to the pressure i7(§ 156); here, however, it is 




Fig. 142. 



DYNAMICS OF A RIGID BODY. 131 

less by hypothesis (perfect rolling). At any instant the four 
unknowns are found by the equations 

2X, i.e., G sin p - P, = (G + g)p ; . (1) 
■2Y,i.e., Gcos/3 - JST, = 0; .... (2) 
2 moms. z , i.e., Pa, = 6Mkz\ • • (3) 

while on account of the perfect rolling, 

6a = p (4) 

Solving, we have, for the acceleration of translation, 

jp = ^ sin /?-[l + (&/-<)]. 
(If the body slid without friction, j? would == ^sin /?.) Hence 
for a cylinder (§ 97), h z % being = -J& 2 , we have^? = %g sin ft ; 
and for a sphere (§ 103) p = \g sin /?. 

(If the plane is so steep or so smooth that both rolling and 
slipping occur, then da no longer =p>, but the ratio of P to -ZV 
is known from experiments on sliding friction ; hence there are 
still four equations.) 

The' motion of translation being thus found to be uniformly 
accelerated, we may use the equations of § 56 for finding dis- 
tance, time, etc. 

Query. — How may we distinguish two spheres by allowing 
them to roll down the same inclined plane, if one of them is 
silver and solid, while the other is of gold, but silvered and 
hollow, so as to be the same as the first in diameter, weight, 
and appearance ? 

125. Parallel-Rod of a Locomotive. — When the locomotive 
moves uniformly, each dM of the rod between the two (or 
three) driving-wheels rotates with ! \ ; 

uniform velocity about a centre of its ** / 2 \ P $ ty I 
own on the line BD, Fig. 143, and \*^~~7 (L) S^^/ 
with a velocity v and radius r common ^_ v 
to all, and likewise has a horizontal ( j.\ ; i At j 
uniform motion of translation. Hence (n.) 

if we inquire what are the reactions P FlG - 143 - 

of its supports, as induced solely by its weight and motion, 
when in its lowest position (independently of any thrust along 



132 MECHANICS OF ENGINEERING. 

the rod), we put 2T of (I.) = 27of (II.) (II. shows the 
imaginary equivalent system), and obtain 

2P - G =fdN =fdMv> + r = (f -~ r)fdM = Mv* -~ r. 

Example. — Let the velocity of translation =50 miles per 
hour, the radius of the pins be 18 in. = f ft., and = half that 
of the driving-wheels, while the weight of the rod is 200 lbs. 
With g = 32.2, we must use the foot and second, and obtain 

v = J[50 X 5280 -i- 3600] ft. per second = 36.6 ; 

while M = 200 -h 32.2 == 200 X .0310 = 6.20 ; 

and finally P = i[200 + 6.2(36.6) 2 -^ f] == 2868.3 lbs., 

or nearly \\ tons, about thirty times that due to the weight 
alone. 

126. So far in this chapter the motion has been prescribed, 
and the necessary conditions determined, to be fulfilled by the 
acting forces at any instant. Problems of a converse nature, 
i.e., where the initial state of the body and the acting forces 
are given while the resulting motion is required, are of much 
greater complexity, but of rare occurrence in practice. The 
reader is referred to Rankine's Applied Mechanics. A treat- 
ment of the Gyroscope will be found in the American Journal 
of Science for 1857, and in the article of that name in Johnson's 
Cyclopaedia. 



WORK, ENERGY, AND POWER. 133 



CHAPTER YI. 

WORK, ENERGY, AND POWER. 

127. Remark. — These quantities as defined and developed 
in this chapter, though compounded of the fundamental ideas 
of matter, force, space, and time, enter into theorems of such 
wide application and practical use as to more than justify their 
■consideration as separate kinds of quantity. 

128. Work in a Uniform Translation. Definition of Work. — 

Let Fig. 144 represent a rigid body having a motion of trans- 
lation parallel to X, acted on by a 
system of forces P„ P„ P 3 , and i? 4 , 
which remain constant. 

Let s be any distance described by 
the bodv during its motion ; then 2 X 
must be zero (§ 109), i.e., noting that 
P i and P 4 have negative X com- 
ponents (the supplements of their 
angles with JT are used), 

P x cos a 1 -\- P 2 cos a 2 — B t cos a z — P 4 cos a K = ; 

or, multiplying by s and transposing, we have (noting that 
s cos a 1 — s x the projection of s on P„ that s cos <x 3 = <? 2 , the 
projection of s on P M and so on), 

A + ^A = ^ + ^ (a) 

The projections s„ s a ; etc., may be called the distances de- 
scribed in their respective directions by the forces P x , P„ ere; 
P 1 and P 2 having moved forward, since s t and s a fall in front 
of the initial position of their points of application ; P z and i? 4 
backward, since s 3 and s A fall behind the initial positions in 
their case. (By forward and backward we refer to the direc- 



*R 3 




) 3 _3*i s ) 




Fig. 144. 



134 MECHANICS OF ENGINEERING. 

tion of each force in turn.) The name Work is given to the 
'product of a force by the distance described in the direction 
of the force by the point of application. If the force moves 
forward (see above), it is called a working force, and is said to 
do the work (e.g., P x s^) expressed by this product ; while if 
backward, it is called a resistance, and is then said to have the 
work (e.g., P s s 3 ), done upon it, in overcoming it through the 
distance mentioned (it might also be said to have done nega- 
tive work). 

Eq. (a) above, then, proves the theorem that : In a uniform 
translation, the working forces do an amount of work which 
is entirely applied to overcoming the resistances. 

129. Unit of Work. — Since the work of a force is a product 
of force by distance, it may logically be expressed as so many 
foot-pounds, inch-pounds, kilogram-meters, according to the 
system of units employed. The ordinary English unit is the 
foot-pound, or ft.-lb. It is of the same quality as a force- 
moment. 

130. Power. — Work as already defined does not depend on 
the time occupied, i.e., the work J D 1 s 1 is the same whether per- 
formed in a long or short time ; but the element of time is of 
so great importance in all the applications of dynamics, as well 
as in such practical commercial matters as water-supply, con- 
sumption of fuel, fatigue of animals, etc., that the rate of work 
is a consideration both of interest and necessity. 

Power is the rate at tohich work is done, and one of its 
units is one foot-pound per second in English practice ; a larger 
one will be mentioned presently. 

The power exerted by a working force, or expended upon a 
resistance, may be expressed symbolically as 

L = P 1 s 1 ~ t, or P 3 s 3 -+- t, 

in which t is the time occupied in doing the work P 1 s l or P 3 s i 
(see Fig. 144) ; or if v x is the component in the direction of 
the force P x of the velocity v of the body, we may also write 

L=P,v (J) 




WORK, ENERGY, AND POWER. 135 

131. Example. — Fig. 145, shows as a free body a sledge 
which is being drawn uniformly up 
a rough inclined plane by a cord 
parallel to the plane. Required the 
total power exerted (and expended), 
if the tension in the cord is P x = 100 ^-^ ^ ^ K3 (WE , GHT) 
lbs., the weight of sledge R 3 = 160 * fig. 145. 

lbs., j3 = 30°, and the sledge moves 240 ft. each minute. JV 
and R A are the normal and parallel (i.e., R A = friction) com- 
ponents of the reaction of the plane on the sledge. From eq. 
(1), § 128, the work done while the sledge advances through 
s = 240 ft. may be obtained either from the working forces, 
which in this case are represented by P x alone, or from the 
resistances R 3 and i? 4 . Take the former method first. Pro- 
jecting s upon P x we have s x = s. 
Hence P.s x or 100 lbs. X 210 ft. = 21,000 ft.-lbs. 
of work done in 60 seconds. That is, the power exerted by the 
working forces is 

L = P x s x -+- t — 400 ft.-lbs. per second. 

As to the other method, we notice that R 3 and i? 4 are resist- 
ances, since the projections s z = s sin /?, and s t = ,9, would fall 
back of their points of application in the initial position, while 
JV is neutral, i.e., is neither a working force nor a resistance, 
since the projection of s upon it is zero. 

From 2X = we have — R, — R 3 sin /3 + P x = 0, 

and from 2 Y = (§ 109) JV - R s cos fi =0; 

whence R % the friction = 20 lbs., and JV — 138.5 lbs. Also, 
since s, = s sin fS = 240 X i = 120 ft., and s 4 = 6', = 240 ft., 
we have for the work done upon the resistances (i.e., in over- 
coming them) in 60 seconds 

R,s % + i?A = 1^0 X 120 + 20 X 240 = 24,000 ft.-lbs., 

and the power expended in overcoming resistances, 

I = 24,000 -r- 60 == 400 ft.-lbs. per second, 

as already derived. Or, in words the power exerted by the 



136 MECHANICS OF ENGINEERING. 

tension in the co v d is expended entirely in raising the weight 
a vertical height of 2 feet, and overcoming the friction through 
a distance of 4 feet along the plane, every second ; the motion 
being a uniform translation. 

132. Horse-Power. — As an average, a horse can exerts a trac- 
tive effort or pull of 100 lbs., at a uniform pace of 4 ft. per sec- 
ond, for ten hours a day without too great fatigue. This gives 
a power of 400 f t.-lbs. per second ; but Boulton & Watt in 
rating their engines, and experimenting with the strong dray- 
horses of London, fixed upon 550 ft.-lbs. per second, or 33,000 
ft.-lbs. per minute, as a convenient large unit of power. (The 
French horse-power, or cheval-vapeur, is slightly less than the 
English, being 75 kilogram meters per second, or 32,550 ft.-lbs. 
per minute.) This value for the horse-power is in common 
use. In the example in § 131, then, the power of 400 ft.-lbs. 
per second exerted in raising the weight and overcoming fric- 
tion may be expressed as (400-^550 =) T 8 T of a horse-power. A 
man can work at a rate equal to about -^ of a horse-powex, 
with proper intervals for eating and sleeping. 

133. Kinetic Energy. Retarded Translation. — In a retarded 

translation of a rigid body whose mass = M, suppose there 

are no working-forces, and that the resistances are constant and 

their resultant is R. (E.g., Fig. 146 shows such a case ; a 

* N sledge, having an initial velocity c and'slid- 

M| j >v ing on a rough horizontal plane, is gradu- 

i^i^^W ' ally retarded by the friction R.) R is par- 

«'q allel to the direction of translation (§ 109) 

fig. 146. and the acceleration is p = — R -f- M; 

hence from vdv —pds we have 

fvdv = - (1 -r- M)/Rds (1) 

But the projection of each ds of the motion upon R is = ds 
itself ; i.e. (§ 128), Rds is the work done upon R, in overcom- 
ing it through the small distance ds, and /Rds is the sum of 
all such amounts of work throughout any definite portion of 
the motion. Let the range of motion be between the points 



WORK, ENERGY, AND POWER. 137 

where the velocity = <?, and where it = zero (i.e., the mass 
has come to rest) With these limits in eq. (1) (0 and s f be- 
ing the corresponding limits for s), we have 

— g- = / Pds (c) 

That is, in giving up all its velocity c the "body has "been able 
to do the work f Pds (this, if P remains constant, reduces to 

Mc 3 

Us') or its equal — o— . If, then, by energy we designate the 

ability to perform work, we give the name kinetic energy of 
a moving body to the product of its mass by half the square 

(Mv\ 
of its velocity \—^-)\ i.e., energy due to motion. (The anti- 
quated term vis viva was once applied to the form Mv 2 .) 

134. Work and Kinetic Energy in any Translation. — Let P 

be the resultant of the working forces at any instant, P that 
of the resistances ; they (§ 109) will both M 

act in a gravity -line parallel to the di- < $ > X 

rection of translation. The acceleration 0«__ s ' *o' 

at any instant is p = (2JE -h M) FlG- 147# 

= (P — P) -r- M\ hence from vdv =pds we have 

Mvdv = Pds — Pds (1) 

Integrating between any two points of the motion as and O f 
where the velocities are v and v\ we have after transposition 






(d) 



But P being the resultant of P 1? P 2 , etc., and P that of 
P ir P 2 , etc., we may prove, as in § 62, that if dn v du 2 , etc., be 
the respective projections of any ds upon P v P 2 > etc., while 
dic\, dw\, etc., are those upon 7? l5 7<? 2 , etc., then 

Pds—P x du x -\-P 2 du 2 -\- .... and Pds=P.dw 1 -\-P 2 dw 2 . . . . ; 

and (d) may be rewritten 



138 MECHANICS OF ENGINEEKING. 

j* P x du, +J* P,du, + . . . . 

/s' ps' VMv' x 3fv s ~l 

Ufa +f a B 2 dw, + . . . . + [_-|- - ^ J ; (e) 

or, in words : In any translation, a portion of the work done 
by the working forces is applied in overcoming the resistances 
ivhile the remainder equals the change in the kinetic energy of 
the body. 

It will be noted that the bracket in (e) depends only on the 
initial and final velocities, and not upon any intermediate 
values ; hence, if the initial state is one of rest, and also the 
final, the total change in kinetic energy is zero, and the work 
of the working forces has been entirely expended in the work 
of overcoming the resistances; but at intermediate stages the 
former exceeds the work so far needed to overcome resistances, 
and this excess is said to be stored in the moving mass ; and as 
the velocity gradually becomes zero, this stored energy becomes 
available for aiding the working forces (which of themselves 
are then insufficient) in overcoming the resistances, and is then 
said to be restored. (The function of a fly-wheel might be 
stated in similar terms, but as that involves rotary motion it 
will be deferred.) 

Work applied in increasing the kinetic energy of a body is 
sometimes called " work of inertia," as also the work done by 
a moving body in overcoming resistances, and thereby losing 
speed. 

135. Example of Steam-Hammer. — Let us apply eq. (e) to 
determine the velocity v' attained by a steam-hammer at the 
lower end of its stroke (the initial velocity being = 0), just 
before delivering its blow upon a forging, supposing that 
the steam-pressure P^ at all stages of the downward stroke is 
given by an indicator. Fig. 148. Weight of moving mass 
is 322 lbs.; .'. H = 10 (foot-pound-second system), 1 = 1 foot. 
The working forces at any instant are P 1 = G = 322 lbs.; P„ 
which is variable, but whose values at the seven equally spaced 



WORK, ENERGY, AND POWER. 



139 



^ZJ 



points a, 5, c, d, e, f, g, are 800, 900, 900, 800, 600, 500, 450 

lbs., respectively. R 1 the exhaust-pressure (16 

lbs. per sq. inch X 20 sq. inches piston-area) = 

320 lbs., is the only resistance, and is constant. 

Hence from eq. (e), since here the projections 

du x , etc., of any ds upon the respective forces 

are equal to each other and = ds, 

pi n i r i Jft/ a 

Pj ds +y P 2 ds = RJ ds + -^-. (1) 

The term fP.ds can be obtained approximately 
by Simpson's Rule, using the above values for 
six equal divisions, which gives 

iV [800 + 4(900 + 800 + 500) 
+ 2(900 + 600) +450] 
= 725 f t.-lbs. of work. Hence, making all the substitutions,, 

we have, since / ds = 1 ft., 



P 

Fig. 148. 



£ 



322 X 1 + 725 = 320 X 1 + iMv n ; .-. \Mv'* = 727ft.-lbs. 
of energy to be expended in the forging. (Energy is evi- 
dently expressed in the same kind of unit as work.) We may 
then say that the forging receives a blow of 727 ft. -lbs. 
energy. The pressure actually felt at the surface of the ham- 
mer varies from instant to instant during the compression of 
the forging and the gradual stopping of the hammer, and 
depends on the readiness with which the hot metal yields. 

If the mean resistance encountered is R m , and the depth of 
compression s", we would have (neglecting the force of gravity, 
and noting that now the initial velocity is v', and the final 
zero), from eq. (c), 

iMv'* = R m s" ; i.e., R m = [727 -f- *" (ft.)] lbs. 

E.g., if s" = | of an inch = ^ of a foot, R m = 43620 lbs., 
and the maximum value of R would probably be about double 
this near the end of the impact. If the anvil also sinks during 
the impact a distance s"\ we must substitute s'" + s" instead 
of s" ; this will give a smaller value for R m . 



140 



MECHANICS OF ENGINEERING. 



By mean value for E is meant [eq. (<?)] that value, E m which 
satisfies the relation 



E m s r =f S Eds. 



This may be called more explicitly a space-average, to dis- 
tinguish it from a time-average, which might appear in some 
problems, viz,, a value E tm , to satisfy the relation (tf being the 
duration of the impact) 

E t J = f fidt, 

and is different from E m . 

From \Mv n = 727 ft.-lbs., we have v' = 12.06 ft. per sec, 
whereas for a free fall it would have been V2 X 32.2x1 = 8.03. 
(This example is virtually of the same kind as Prob. 4, § 59, 
differing chiefly in phraseology.) 



136. Pile-Driving. — The safe load to be placed upon a pile after 
the driving is finished is generally taken as a fraction (from -§- 
to -§) of the resistance of the earth to the passage of the pile as 
indicated by the effect of the last few blows of the ram, in ac- 
cordance with the following approximate theory : Toward the 
t end of the driving the resistance E encountered by 

the pile is nearly constant, and is assumed to be that 
J met by the ram at the head of the pile; the distance 

s' through which the head of the pile sinks as an 
Mil effect of the last blow is observed. If G, then, is 

the weight of the ram, = Mg, and h the height of 
free fall, the velocity due to A, on striking the pile, 
isc= V%gh (§ 52), and we have, from eq. (<?), 

\Me\ i.e., Gh, = f Eds = Es' . . (1) 

(E being considered constant) ; hence E = Gh -r- s'. 
and the safe load (for ordinary wooden piles), 

P = from | to i of Gh ^ s' (2) 

Maj. Sanders recommends \ from experiments made at Fort 



Fig. 149. 



WOEK, ENEKGY, AND POWER. 141 

Delaware in 1851; Molesworth, J-; General Barnard, -J-, from 
extensive experiments made in Holland. 

Of course from eq. (2), given jP, we can compute s f . 

(Owing to the uncertainty as to how much of the resistance 
i? is due to friction of the soil on the sides of the pile, and 
how much to the inertia of the soil around the shoe, the more 
elaborate theories of Weisbach and Rankine seem of little 
practical account.) 

137. Example. — In preparing the foundation of a bridge-pier 
it is found that each pile (placing them 4 ft. apart) must bear 
safely a load of 72 tons. If the ram weighs one ton, and falls 
12 ft., what should be the effect of the last blow on each pile? 
Using the foot- ton-second system of units, and Molesworth 
factor •£-, eq. (2) gives 

s' = i(l X 12 + 72) = ^ of a foot = J of an inch. 

That is, the pile should be driven until it sinks only -J inch 
under each of the last few blows. 

138. Kinetic Energy Lost in Inelastic Direct Central Impact.— 

Referring to § 60, and using the same notation as there given, 
we find that if the united kinetic energy possessed by two in- 
elastic bodies after their impact, viz., ^Mfi* + i^C 2 , Shav- 
ing the value (M x c x + M 2 c a ) -f- (M 1 + M 9 ), be deducted from 
the amount before impact, viz., iM^* + \M^c^ the loss of 
Tcinetic energy during impact of two inelastic bodies is 

w ~ m\ + MP c *> W 

An equal amount of energy is also lost by partially elastic 
bodies during the first period of the impact, but is partly re- 
gained in the second. If the bodies were perfectly elastic, we 
would find it wholly regained and the resultant loss zero, from 
the equations of § 60 ; but this is not quite the reality, on 
account of internal vibrations. 

The kinetic energy still remaining in two inelastic bodies 
after impact (they move together as one mass) is 



142 



MECHANICS OF ENGINEERING. 



\{M X -\- M^) C*, or, after inserting the value of 
G = (M x c x + M&) -r- (M x + if 2 ), we have 

1 [M x c x +M 2 cJ 



W = 



M x + M, 



(2) 



Example 1. — The weight G x = M x g falls freely 
through a height A, impinging upon a weight 6r 9 
= M 3 g, which was initially at rest. After their (in- 
elastic) impact they move on together with the com- 
bined kinetic energy just given in (2), which, since 
c x and c a , the velocities before impact, are respectively 
V2gh and 0, may be reduced to a simpler form. 
This energy is soon absorbed in overcoming the 
flange-pressure R, which is proportional (so long as 
the elasticity of the rod is not impaired) to the 
elongation s, as with an ordinary spring. If from 
previous experiment it is known that a force R 

produces an elongation s , then the variable R = (.R -=- s Q )s. 

Neglecting the weight of the two bodies as a working force, 

we now have, from eq. (d), 

s n 



cz> 



els 



Fig. 150. 







R n 



i.e.. 



°- f sds + 
M;gh 



M x + M, 



M x +M, 



(3) 



"When s = s\ i.e., when the masses are (momentarily) at rest 
in the lowest position, the flange-pressure or tensile stress in the 
rod is a maximum, R' — (R -f- s )s f , whence s' = R f s Q -f- R ? ; 
and (3) may be written 



2 

R'X 



s' ==■ 



M*gh 

m x +m; 

M?gh 



or 



2R ' M x + M, 



(4) 

(5) 



Eq. (3) gives the final elongation of the rod, and (5) the greatest 
tensile force upon it, provided the elasticity of the rod is not 



WORK, ENERGY, AND POWER. 143 

impaired. The form f It's' in (4) may be looked upon as a direct 

integration of / fids, viz., the mean resistance tyR') multi- 

plied by the whole distance (V) gives the work done in over- 
coming the variable R through the successive ds\. 

If the elongation is considerable, the working-forces G 1 and 
G^ cannot be neglected, and would appear in the term -}-(#, 
+ G 2 )s' in the right-hand members of (3) , (4), and (5). The 
upper end of the rod is firmly fixed, and the rod itself is of 
small mass compared with M 1 and Jf 2 . 

Example 2. — Two cars, Fig. 151, are connected by an elastic 
chain on a horizontal track. Velocities before impact (i.e., 

before the stretching of the chain be- >c 2 >-ci 

gins, by means of which they are 



brought to a common velocity at the " M 9 M a 

instant of greatest tension R ', and fig. 151. 

elongation s / of the chain) are c x = c t , and c 2 = 0. 

During the stretching, i.e., the first period of the impact, the 
kinetic energy lost by the masses has been expended in stretch- 
ing the chain, i.e., in doing the work \R's' ; hence we may 
write (the elasticity of the chain not being impaired) (see eq. (1) ) 

in which the different symbols have the same meaning as in 
Example 1, in which the rod corresponds to the chain of this 
example. 

In this case the mutual accommodation of velocities is due 
to the presence of the chain, whose stretching corresponds to 
the compression (of the parts in contact) in an ordinary impact. 

In numerical substitution, 32.2 for g requires the use of the 
units foot and second for space and time, while the unit of 
force may be anything convenient. 

139. Work and Energy in Rotary Motion. Axis Fixed. — 

The rigid body being considered free, let an axis through 
perpendicular to the paper be the axis of rotation, and resolve 
all forces not intersecting the axis into components parallel 



144 



MECHANICS OF ENGINEERING. 



and perpendicular to the axis, and the latter again into com- 
ponents tangent and normal to the circular path of the point 

of application. These tangential com* 
ponents are evidently the only ones 
of the three sets mentioned which 
have moments about the axis, those 
having moments of the same sign as 
go (the angular velocity at any instant) 
being called working forces, T x , T 2 , 
etc. ; those of opposite sign, resist- 
ances, T/, T 2 ', etc.; for when in time 
dt the point of application B x , of T x , describes the small arc 
ds 1 = a x da, whose projection on T 1 is = ds x , this projection 
falls ahead (i.e., in direction of force) of the position of the 
point at the beginning of dt, while the reverse is true for T t \ 




Fig. 152. 



From eq. (XIV.), § 114, we have for 6 (angul. accel.) 



e 



i 



(i> 



which substituted in oodoo = Oda (from § 110) gives (remem- 
bering that a^doc = ds x , etc.), after integration and transposition,. 

T^+J TA + etc. 

=£ T;a$; +f Tjds; + etc. + [£</ - £</], (2) 

where and n refer to any two (initial and final) positions of 
the rotating body. Eq. (4), § 120, is an example of this. 

JSTow \oo^I— \oo*fdMf> % =fidM(GD n py, which, since oo n f> 
is the actual velocity of any dM &t this (final) instant, is nothing 
more than the sum of the amounts of kinetic energy possessed 
at this instant by all the particles of the body ; a similar state- 
ment may be made for ioo 2 I. 

Eq. (2) therefore may be put into words as follows: 

Between any two positions of a rigid body rotating about a 

fixed axis, the work done by the working forces is partly used 

in overcoming the resistances, and the remainder in changing 

the kinetic energy of the individual particles. If in any case 



WORK, ENERGY, AND POWER. 145 

this remainder is negative, the final kinetic energy is less than 
the initial, i.e., the work done by the working forces is less than 
that necessary to overcome the resistances through their respec- 
tive spaces, and the deficiency is made up by the restoring of 
some of the initial kinetic energy of the rotating body. A 
moving fly-wheel, then, is a reservoir of kinetic energy. 

Eq. (2) has already been illustrated numerically in § 121, 
where the additional relation was utilized (for a connecting-rod 
and piston of small mass), that the work done in the steam- 
cylinder is the same as that done directly at the crank-pin by 
the working-force there. 

140. Work of Equivalent Systems the Same. — If two plane 

systems of forces acting on a rigid body are equivalent (§ 1 5 a), 
the aggregate work done by either of them during a given slight 
displacement or motion of the body parallel to their plane is 
the same. By aggregate work is meant what has already been 
defined as the sum of the " virtual moments" (§§ 61 to 64), in 
any small displacement of the body, viz., the algebraic sum of 
the products, 2 (Pdu), obtained by multiplying each force by 
the projection (du) of the displacement of (or small space 
described by) its point of application upon the force. (We 
here class resistances as negative working forces.) 

Call the systems A and B ; then, if all the forces of B were 
reversed in direction and applied to the body along with those 
of A. the compound system would be a balanced system, and 
hence we would have (§ 64), for a small motion parallel to the 
plane of the forces, 

2(Pdu) = 0, i.e., 2(Pdu) for A - 2(Pdu) for B = 0, 

or -f 2(Pdu) for A = + 2(Pdu) for B. 

But -f- 2 (Pdu) for A is the aggregate work done by the forces 
of A during the given motion, and -\- 2 (Pdu) for B is a 
similar quantity for the forces of B (not reversed) during the 
same small motion if B acted alone. Hence the theorem is 
proved, and could easily be extended to space of three dimen- 
sions. 

10 



146 



MECHANICS OF ENGINEERING. 




141. Relation of Work and Kinetic Energy for any Extended 
Motion of a Rigid Body Parallel to a Plane. — (If at any instant 

any of the forces acting are not 
parallel to the plane mentioned, 
their components lying in or 
parallel to that plane, will be used 
instead, since the other compo- 
nents obviously would be neither 
working forces nor resistances.) 
Fig. 153 shows an initial position, 
and any intermediate, as q. The 
forces of the system acting may vary in any manner during 
the motion. 

In this motion each dM describes a curve of its own with 
varying velocity v, tangential acceleration j> t , and radius of 
curvature r ; hence in any position q, an imaginary system B 
(see Fig. 154), equivalent to the actual system A (at q in Fig. 
153), would be formed by applying to each dM a ~| 



Fig. 153. 

of the body; a 



final, 



n 



tangential force dT = dMp t , and a normal force , 
dN = dMv 2 -=- r. By an infinite number of con- 
secutive small displacements, the body passes from 
o to n. In the small displacement of which q is the 
initial position, each dM describes a space ds, and FlG . 154. 
d T does the work dTds = dMvdv, while dN does the work- 
dN X0 = 0. Hence the total work done by B in the small 
displacement at q would be 



= dM'v'dv' + dM"v"dv" + etc., 



(1) 



including all the dM 's of the body and their respective veloci- 
ties at this instant. 

But the work at q in Fig. 153 by the actual forces (i.e., of 
system A) during the same small displacement must (by § 140) 
be equal to that done by B. hence 

P x du x + P 7 du, + etc. s= dM'v'dv' + dM"v"dv" + etc. (q) 

Now conceive an equation like (q) written out for each of 



WORK, ENERGY, AND POWER. 147 

the small consecutive displacements between positions o and 
n and corresponding terms to be added ; this will give 

Pjlu x -\- I P a du 9 + etc. 

= dM'j^v'W + dM' f f n v"dv" + etc. 

= yw\vS - O + idlT'W* - v ">) + etc. 
The second member may be rewritten so as to give, finally, 

P.du.+j P^+etc. = 2QdMv n *) -2{idMv 2 ), (XY.) 

or, in words, the work done by the acting forces {treating a re- 
sistance as a negative working force) between any two posi- 
tions is egual to the gain {or loss) in the aggregate kinetic 
energy of the particles of the body between the two positions. 
To avoid confusion, 2 has been used instead of the signy" in 
one member of (XY.), in which v n is the final velocity of any 
dM (not the same for all necessarily) and v the initial. 

(The same method of proof can be extended to three dimen- 
sions.) 

Since kinetic energy is always essentially positive, if an ex- 
pression for it comes out negative as the solution of a problem, 
some impossible conditions have been imposed. 

142. Work and Kinetic Energy in a Moving Machine. — 

Defining a mechanism or machine as a series of rigid bodies 
jointed or connected together, so that working-forces applied 
to one or more may be the means of overcoming resistances 
occurring anywhere in the system, and also of changing the 
amount of kinetic energy of the moving masses, let us for 
simplicity consider a machine the motions of whose parts are 
all parallel to a plane, and let all the forces acting on any one 
piece, considered free, at any instant be parallel to the same 
plane. 

Now consider each piece of the machine, or of any series of 
its pieces, as a free body, and write out eq. (XY.) for it be- 
tween any two positions (whatever initial and final positions are 



148 



MECHANICS OF ENGINEERING. 



selected for the first piece, those of the others must be corre- 
sponding initial and corresponding final positions), and it will 
be found, on adding up corresponding members of these equa- 
tions, that the terms involving those components of the mutual 
pressures (between the pieces considered) which are normal 
to the rubbing surfaces at any instant w r ill cancel out, while 
their components tangential to the rubbing surfaces (i.e., fric- 
tion, since if the surfaces are perfectly smooth there can be 
no tangential action) will appear in the algebraic addition as 
resistances multiplied by the distances rubbed through, meas- 
ured on the rubbing surfaces. For example, Fig. 155, where 
one rotating piece both presses and rubs on another. Let the 
normal pressure between them at A be R 2 == P t ; it is a work- 
ing force for the body of mass M" , but a resistance for M' , 
hence the separate symbols for the numerically equal forces 
(action and reaction). 

Similarly, the friction at A is i? 3 = _P S ; a resistance for M\ 
a working-force for M" . (In some cases, of course, friction 
may be a resistance for both bodies.) For a small motion, A 
describes the small arc AA' about 0' in dealing with M\ but 
for M" it describes the arc AA" about 0' ', A' A" being 
parallel to the surface of contact AD, while AB is perpen- 




Fig. 155. 



Fig. 156. 



Fig. 157. 



dicular to A'A' r . In Figs. 156 and 157 we see W and M" 
free, and their corresponding small rotations indicated. During 
these motions the kinetic energy (K. E.) of each mass has 
changed by amounts d(K. E.) M / and d(K. ~E.)m» respectively, and 
hence eq. (XV.) gives, for each free body in turn, 

P'a'a 1 - R,AB - R,TB = d(K. E.) M > . (1) 

- RW + P.AB + P^^B = d(K. E. V. . (2) 



WORK, ENERGY, AND POWER. 149 

Now add (1) and (2), member to member, remembering that 
P 2 = P 2 and P 3 = P z = F 3 = friction, and we have 



P x aa' - F.A'A" - R]jb" = d(K. E.) M , + d(K. E. V, (3) 

in which the mutual actions of M 7 and M" do not appear, 
except the friction, the work done in overcoming which, when 
the two bodies are thus considered collectively, is the product 
of the friction by the distance A' A" of actual rubbing meas- 
ured on the rubbing surface. For any number of pieces, then, 
considered free collectively, the assertion made at the beginning 
of this article is true, since any finite motion consists of an 
infinite number of small motions to each one of which an equa- 
tion like (3) is applicable. 

Summing the corresponding terms of all such equations, we 
have 

f U P,du x J r J ,n p 2 du,+ etc. = 2(KE.)n- ^(K.E.) .(XVI.) 

This is of the same form as (XV.), but instead of applying to a 
single rigid body, deals with any assemblage of rigid parts 
forming a machine, or any part of a machine (a similar proof 
will apply to three dimensions of space); but it must be remem- 
bered that it excludes all the mutual actions of the pieces con- 
sidered except friction, which is to be introduced in the manner 
just illustrated. A flexible inextensible cord may be considered 
as made up of a great number of short rigid bodies jointed 
without friction, and hence may form part of a machine with- 
out vitiating the truth of (XVI.). 

^(K. E.) n signifies the sum obtained by adding the amounts 
of kinetic energy (%dMv n * for each elementary mass) possessed 
by all the particles of all the rigid bodies at their final posi- 
tions ; ^(K. E.) , a similar sum at their initial positions. For 
example, the K. E. of a rigid body having a motion of transla- 
tion of velocity v, = \vfdM = %Mv 2 ; that of a rigid body 
having an angular velocity go about a fixed axis Z, = \<^Iz 
(§ 139) ; while, if it has an angular velocity go about a gravity- 



150 MECHANICS OF ENGINEERING. 

axis Z, which has a velocity v z of translation at right angles to 
itself, the (K. E.) at this instant may be proved to be 

i.e., is the sum of the amounts due to the two motions sepa- 
rately. 

143. K. E. of Combined Rotation and Translation. — The last 
statement may be thus proved. Fig. 158. 
At a given instant the velocity of anv dM is 

up ° J J 

V v, the diagonal formed on the velocity v z of 



-<a /§M\ Vz translation, and the rotary velocity oop rela- 
^ / \ tively to the moving gravity-axis Z (per- 

/ \ pendicular to paper) (see § 71), 



Fig. 158. ^ ^ — ®Z + Op)* ~ ^(pop)v z COS cp ; 

hence we have K. E., at this instant, 

= fidMv* =iv z i fdM + WfdM-P' 2 - ooVzfdMp cos q>, 

but p cos q> = y, and fdMy = My = 0, since Z is a gravity- 
axis, 

.-. K. E. = \Mv z * + %gd*I z . Q- E. D. 

It is interesting to notice that the K. E. due to rotation, viz., 
\o?I z = \M(ooltf, is the same as if the whole mass were con- 
centrated in a point, line, or thin shell, at a distance &, the 
radius of gyration, from the axis. 

144. Example of a Machine in Operation. — Fig. 159. Con- 
sider the four consecutive moving masses, M\ M", JS/ 7 ", and 
M iv (being the piston ; connecting-rod ; fly-wheel, crank, drum, 
and' chain ; and weight on inclined plane) as free, collectively. 
Let us apply eq. (XYL), the initial and final positions being 
taken when the crank-pin is at its dead-points o and n\ i.e., we 
deal with the progress of the pieces made while the crank-pin 
describes its upper semicircle. Remembering that the mutual 
actions between any two of these four masses can be left out 
of account (except friction), the only forces to be put in are 
the actions of other bodies on each one of these four, and are 



WORK, ENERGY, AND POWER. 



151 



shown in the figure. The only mutual friction considered will 
be at the crank- pin, and if this as an average = F\ the work 
done on it between o and n = F'nr" , where r" = radius of 
crank-pin. The work done by P 1 the effective steam- pressure 
(let it be constant) during this period is = PJ,' ; that done in 
overcoming F 19 the friction between piston and cylinder, = FJJ ; 
that done ujpon the weight G r oi connecting-rod is cancelled by 
the work done by it in the descent following ; the work done 




Fig. 159. 



upon G iv , = G Xv 7ta sin /?, where a = radius of drum ; that 
upon the friction F A , = Fjra. The pressures JV, N', JV ]y , and 
N'" , and weights G' and G"\ are neutral, i.e., do no work either 
positive or negative. Hence the left-hand member of (XYI.) 
becomes, between o and n, 

p$ - F X V - F' r 7tr" - G u na sin fi - Fjta, . . (1) 

provided the respective distances are actually described by 
these forces, i.e., if the masses have sufficient initial kinetic 
energy to carry the crank-pin beyond the point of minimum 
velocity, with the aid of the working force P„ whose effect is 
small up to that instant. 

As for the total initial kinetic energy, i.e., ^(K. E.) , let us 
express it in terms of the velocity of crank-pin at o, viz., V . 
The (KE.) of M' is nothing ; that of M", which at this in- 
stant is rotating about its right extremity {fixed for the instant) 
with angular velocity go" = T -f- l'\ is \od"' 1 W'\ that of M'" 
= %Go nn I c "\ in which go"' = V -±-r; that of M iv (translation) 
= i M iv v "% in which v lv = {a-~r) V . 2(K. E.)» is expressed 



152 



MECHANICS OF ENGINEERING. 



in a corresponding maimer with Y n (final velocity of crank-pin) 
instead of V . Hence the right-hand member of (XYI.) will 
give (putting the radius of gyration of M." about 0" == k'\ 
and that of M" about C=Ib) 



i(Vn - vd\ji;^+w'£+M»pJ 



(*) 



By writing (1) = (2), we have an equation of condition, capa- 
ble of solution for any one unknown quantity, to be satisfied 
for the extent of motion considered. It is understood that the 
chain is always taut, and that its weight and mass are neg- 
lected. 



145. Numerical Case of the Foregoing. — (Foot-pound-second 
system of units for space, force, and time ; this requires g 
= 32.2.) 

Suppose the following data : 



Feet. 


Lbs. 


Lbs. 


Mass Units. 


V = 2.0 
I" = 4.0 
a = 1.5 
r = 1.0 

k = 1.8 
k" = 2.3 
r" = 0.1 


Pi = 

Fx = 

F" (av'ge) = 

F± = 


6000 
200 
400 
300 


G' = 60 
G" = 50 
G'" = 400 
G" = 3220 


(aud . \) 

M' = 1.86 
M" = 1.55 
M"> = 12.4 
J/ iv = 100.0 


Also let V = 4 


ft. per sec.; /S=30° 



Denote (1) by IF and the large bracket in (2) by M (this by 
some is called the total mass u reduced" to the crank-pin). 
Putting (1) = (2) we have, solving for the unknown V n , 



F.= 



2 IF 



M 



V'. 



(3) 



For above values, 

W = 12,000 - 400 - 125.7 - 7590.0 - 1417.3 
= 2467 foot-pounds ; 
while M = 0.5 + 40.3 + 225.0 = 265.8 mass-units; 
whence 



V n = 1/18.56 + 16 = ^34.56 =± 5.S8 ft. per second. 



153 

As to whether the crank-pin actually reaches the dead-point 
n, requires separate investigations to see whether V becomes 
zero or negative between o and n (a negative value is inad- 
missible, since a reversal of direction implies a different 
value for IF), i.e., whether the proposed extent of motion is 
realized ; and these are made by assigning some other inter- 
mediate position m, as a final one, and computing Y m . remem- 
bering that when m is not a dead-point the (K. E.) m of W is not 
zero, and must be expressed in terms of Y m , and that the 
(K. E.) m of the connecting-rod M" must be obtained from § 143 

146. Regulation of Machines. — As already illustrated in 
several examples (§ 121), a fly-wheel of sufficient weight and 
radius may prevent too great fluctuation of speed in a single 
stroke of an engine ; but to prevent a permanent change, which 
must occur if the work of the working force or forces (such as 
the steam-pressure on a piston, or water-impulse in a turbine) 
exceeds for several successive strokes or revolutions the work 
required to overcome resistances (such as friction, gravity, re- 
sistance at the teeth of saws, etc., etc.) through their respective 
spaces, automatic governors are employed to diminish the 
working force, or the distance through which it acts per stroke, 
until the normal speed is restored ; or vice versa, if the speed 
slackens, as when new resistances are temporarily brought into 
play. Hence when several successive periods, strokes (or other 
cycle), are considered, the kinetic energy of the moving parts 
will disappear from eq. (XVI.), leaving it in this form : 

work of working -forces = work done itpon resistances. 

147. Power of Motors. — In a mill where the same number of 
machines are run continuously at a constant speed proper for 
their work, turning out per hour the same number of barrels 
of flour, feet of lumber, or other commodity, the motor (e.g., 
a steam-engine, or turbine) works at a constant rate, i.e., de- 
velops a definite horse-power (H.P.), which is thus found in 
the case of steam-engines (double-acting) : 



154 MECHANICS OF ENGINEERING. 

H.P. = total mean effective ) / distance in feet ) 

steam-pressure on > x| travelled by pis- > ~ 550, 
piston in lbs. ) ( ton per second. ) 

i.e., the work (in ft.-lbs) done per second by the working force 
divided by 550 (see § 132). The total effective pressure at any 
instant is the excess of the forward over the back-pressure, 
and by its mean value (since steam is usually used expansively) 
is meant such a value P' as, multiplied by the length of stroke 
I, shall give 

P'l=fPdx, 

where P is the variable effective pressure and dx an element 
of its path. If u is the number of strokes per second, we may 
also write {foot-pound- second system) 



H.P. = P'lu -r- 550 



/ 



Pdx 



u + 550. (XVII.) 



Very often the number of revolutions per minute, m, of the 
crank is given, and then 

H.P. = P' (lbs.) X 2Z (feet) X m -f- 33,000. 

If P= area of piston we may also write P' — Fp f , where p' 
is the mean effective steam-pressure per unit of area. Evi- 
dently, to obtain P' in lbs., we multiply J^in sq. in. hjp f in 
lbs. per sq. in., or F in sq. ft. by p' in lbs. per sq. foot ; the 
former is customary, p' in practice is obtained by measure- 
ments and computations from " indicator-cards" (see § 135, in 
which (P 2 — P x ) corresponds to P of this section) ; or P% i.e., 

/ Pdx, may be computed theoretically as in § 59, Problem 4. 

The power as thus found is expended in overcoming the 
friction of all moving parts (which is sometimes a large item), 
and the resistances peculiar to the kind of work done by the ma- 
chines. The work periodically stored in the increased kinetic 
energy of the moving masses is restored as they periodically 
resume their minimum velocities. 



AND POWER. 155 

148. Potential Energy. — There are other ways in which work 
or energy is stored and then restored, as follows : 

First. In raising a weight G through a height h, an amount 
of work = Gh is. done upon G, as a resistance, and if at any 
subsequent time the weight is allowed to descend through the 
same vertical distance h (the form of path is of no account), G y 
now a working force, does the work Gh, and thus in aiding the 
motor repays, or restores, the Gh expended by the motor in 
raising it. If h is the vertical height through which the centre 
of gravity rises and sinks periodically in the motion of the 
machine, the force G may be left out of account in reckoning 
the expenditure of the motor's work, and the body when at its 
highest point is said to possess an amount Gh of potential 
energy, i.e., energy of position, since it is capable of doing the 
work Gh in sinking through its vertical range of motion. 

Second. So far, all bodies considered have been by express 
stipulation rigid, i.e., incapable of changing shape, To see 
the effect of a lack of rigidity as affecting the principle of 
work and energy in machines, ^^^sC^^rirr^- 

take the simple case in Fig. 160. p jT^^^U^ £ 

A helical spring at a given in- ^f-«^^^4^^ 
Btant is acted on at each end by f&M g hd& v 

a force P in an axial direction ' "' j / 

(they are equal, supposing the fig. ieo. 

mass of the spring small). As the machine operates of which 
it is a member, it moves to a new consecutive position B, 
suffering a further elongation dX in its length (if P is increas- 
ing). P on the right, a working force, does the work Pdx'', 
how is this expended ? P on the left has the work Pdx done 
upon it, and the mass is too small to absorb kinetic energy or 
to bring its weight into consideration. The remainder, Pdx' 
— Pdx = PdX, is expended in stretching the spring an addi- 
tional amount d\ and is capable of restoration if the spring 
retains its elasticity. Hence the work done in changing the 
form of bodies if they are elastic is said to be stored in the 
form of potential energy. That is, in the operation of ma- 
chines, the name potential energy is also given to the energy 



156 MECHANICS OF ENGINEERING. 

stored and restored periodically in the changing and regaining 
of form of elastic bodies. 

149. Other Forms of Energy. — Numerous experiments with 
various kinds of apparatus have proved that for every 772 
(about) ft.-lbs. of work spent in overcoming friction, one British 
unit of heat is produced (viz., the quantity of heat necessary to 
raise the temperature of one pound of water from 32° to 33° 
Fahrenheit); while from converse experiments, in which the 
amount of heat used in operating a steam-engine was all carefully 
estimated, the disappearance of a certain portion of it could only 
be accounted for by assuming that it had been converted into 
work at the same rate of (about) 772 ft.-lbs. of work to each 
unit of heat (or 425 kilogrammetres to each French unit of 
heat). This number 772, or 425, according to the system of 
units employed, is called the Mechanical Equivalent of Heat, 
first discovered by Joule and confirmed by Hirn.* 

Heat then is energy, and is supposed to be of the kinetic 
form due to the rapid motion or vibration of the molecules of 
a substance. A similar agitation among the molecules of the 
(hypothetical) ether diffused through space is supposed to pro- 
duce the phenomena of light, electricity, and magnetism. 
Chemical action being also considered a method of transform- 
ing energy (its possible future occurrence as in the case of coal 
and oxygen being called potential energy), the well-known 
-doctrine of the Conservation of Energy, in accordance with 
which energy is indestructible, and the doing of work is simply 
the conversion of one or more kinds of energy into equivalent 
amounts of others, is now one of the accepted hypotheses of 
physics. 

Work consumed in friction, though practically lost, still re- 
mains in the universe as heat, electricity, or some other subtile 
form of energy. 

150. Power Required for Individual Machines. Dynamome- 
ters of Transmission. — If a machine is driven by an endless 
belt from the main-shaft, A, Fig. 161, being the driving-pulley 

* Prof. Rowland's recent experiments result in the value 429.8 kilogram- 
metres at a temperature of 5° Cent. 



WORK, ENERGY, AND POWER. 



15? 



on the machine, the working force which drives the machine, 
in other words the " grip" with which the 
belt takes hold of the pulley tangentially, 
= P — P\ P and P' being the tensions 
in the "driving" and "following" sides of 
the belt respectively. The belt is supposed 
not to slip on the pulley. If v is the ve- 
locity of the pulley -circumference, the fig. iei. 
work expended on the machine per second, i.e., the power, is 




L = (P - P>. 



(1) 



To measure the force (P — P f ), an apparatus called a Dy- 
namometer of Transmission maybe placed between the main 
shaft and the machine, and the belt made to pass through it in 
such a way as to measure the tensions P and P' , or princi- 
pally their difference, without meeting any resistance in so do- 
ing ; that is, the power is transmitted, not absorbed, by the 
apparatus. One invention for this purpose (mentioned in the 
Journal of the Franklin Institute some years ago) is shown 

{in principle) in Fig. 162. A ver- 
tical plate carrying four pulleys and 
a scale-pan is first balanced on the 
pivot G. The belt being then ad- 
justed, as shown, and the power 
turned on, a sufficient weight G is 
placed in the scale-pan to balance 
the plate again, for whose equilib- 
rium we must have Gb = Pa — P'a, since the P and P' on 
the right are purposely given no leverage about C. The ve- 
locity of belt, v, is obtained by a simple counting device. 
Hence (P — P r ) and v become known, and .*. L from (1). 

Many other forms of transmission-dynamometers are in use, 
some applicable whether the machine is driven by belting or 
gearing from the main shaft. Emerson's Hydrodynamics de- 
scribes his own invention on p. 283, and gives results of meas- 
urements with it ; e.g., at Lowell, Mass., the power required 
to drive 112 looms, weaving 36-inch sheetings, No. 20 yarn, 




158 MECHANICS OF ENGINEERING. 

60 threads to the inch, speed 130 picks to the minute, was 
found to be 16 H.P., i.e., \ H.P. to each loom (p. 335). 

151. Dynamometers of Absorption. — These are so named 
since they furnish in themselves the resistance (friction or a 
weight) in the overcoming (or raising) of which the power is 
expended or absorbed. Of these the Prony Friction Brake 
is the most common, and is used for measuring the power 
developed bj a given motor (e.g., a steam-engine or turbine) 
not absorbed in the friction of the motor itself. Fig. 163 



Fig. 163. 

shows one fitted to a vertical pulley driven by the motor. By 
tightening the bolt J5, the velocity v of pulley-rim may be 
made constant at any desired value (within certain limits) by 
the consequent friction, v is measured by a counting appara- 
tus, while the friction (or tangential components of action be- 
tween pulley and brake), = F, becomes known by noting the 
weight G which must be placed in the scale pan to balance the 
arm between the checks ; then 

Fa=Gh, (1) 

for the equilibrium of the brake (supposing the weight of 
brake and scale-pan previously balanced on C) and the work 
done per unit of time, or power, is 

Z = Fv (2) 

A " dash-pot " is frequently connected with the arm to prevent 
sudden oscillations. In case the pulley is horizontal, a bell- 
crank lever is added between the arm and the scale-pan, and 
then eq. (1) will contain two additional lever-arms. 



WORK, ENERGY, AND POWER. 



159 



Po 


Pi 


p 2 


p 3 


b 

P 4 


i 

i 

P 5 ^N 














P 6 ) 



Fig. 164. 



152. The Indicator, used with steam and other fluid engines, 
is a special kind of dynamometer in which the automatic mo- 
tion of a pencil describes a curve 
on paper whose ordinates are 
proportional to the fluid pres- 
sures exerted in the cylinder at 
successive points of the stroke. 
Thus, Fig. 164, the back-pres- 
sure being constant and — P b , 
the ordinates P , JP 19 etc., represent the effective pressures at 
equally spaced points of division. The mean effective pressure 
P' (see § 147) is, for this figure, by Simpson's Rule (six equal 
spaces), 

P" = tVCA + HP, + P, + P>) + 2CA + P t ) + P.!- 

This gives a near approximation. The power is now found by 
§147. 

153. The theory of Atwood's Machine is most directly ex- 
pressed by the principle of work and energy ; i.e., by eq. 

(XYL), §142. Fig. 165. The parts 
considered free, collectively, are the 
rigid bodies P, Q, G, and four friction- 
wheels like G x \ and the flexible cord, 
which does not slip on the upper pul- 
ley. There is no slipping at P, hence 
no sliding friction there. The actions 
of external bodies on these eight consist 
of the working force P, the resistances 
Q and the four jP's (at bearings of fric- 
tion-wheel axles); all others (G, 4:G^ 
and the four i?'s) are neutral. Since there is no rubbing be- 
tween any two of the eight bodies, no mutual actions whatever 
will enter the equation. Let P > Q, and Zand I x be the mo- 
ments of inertia of G and G„ respectively, about their respec- 
tive axes of figure. Let the apparatus start from rest, then 
when P has descended through any vertical distance s, and ac- 




Fig. 165. 



160 MECHANICS OF ENGINEERING. 

quired the velocity v, Q has been drawn up an equal distance 
and acquired the same velocity, while the pulley G has ac- 
quired an angular velocity gd = v -f- a, each friction-pulley an 
angular velocity gd 1 = (r : a)v 4- a v As to the forces, P Las- 
done tLe work Ps, Q has had the work Qs done upon it, while 
each i^has been overcome through the space (r r : a^){r : a)s; 
all the other forces are neutral. Hence, from eq. (XVL), § 142 
(see also § 139), we have 

sp /p 

Ps-Qs - 4F- 1 . -s 

^ a x a 

~Lg ^ gJ2 ^ 2 a 2± ^ 2 * a 2 ' a 2 ' 

Evidently «=^x constant, i.e., the motion of JP and Q is 
uniformly accelerated. If, after the observed space s has been 
described, P is suddenly diminished to such a value P' that 
the motion continues with a constant velocity = v, we shall 
have, for any further space s', 

^ a x a 

from which i^can be obtained (nearly) ; while if if be the ob- 
served time of describing s\ v = s' -j- t' becomes known. 
Also we may write 1= (G -f- g)lc 2 and I x = (6^ -f- (Z)^ 2 , and 
thus finally compute the acceleration of gravity, g, from our 
first equation above. 

154. Boat-Rowing.— ^'ig. 166. During the stroke proper, 
let P = mean pressure on one oar-handle ; hence the pressures 
on the foot-rest are 2P, resistances. Let Jf=massof boat 
and load, w and v n its velocities at beginning and end of stroke. 
P x = pressures between oar-blade and water. P = mean re- 
sistance of water to the boat's passage at this (mean) speed. 
These are the only (horizontal) forces to be considered as act- 
ing on the boat and two oars, considered free collectively. 
During the stroke the boat describes the space s 3 = OP, the 
oar-handle the space s 2 = AB, while the oar-blade slips back- 



WORK, ENERGY, AND POWER. 



161 



ward through the small space (the " slip") = s l (average). 
Hence by eq. (XYI.), § 142, 

2Ps, - 2Ps 3 -P s 3 - 2P A = \Miy: - <) ; 
i.e., 2P{s i -s z ) = 2PxTE=2Ps =Ps 3 +2P 1 s 1 + |Jf«-<); 

or, in words, the product of the oar-handle pressures into the 
distance described by them measured on the boat, i.e., the work 
done by these pressures relatively to the boat, is entirely ac- 
counted for in the work of slip and of liquid-resistance, and in- 



'Pi 



-VD 



F 


S\ r 


u 


?):F 


VbN 


R 






















^■^ 




v P/ 





Fig. 166. 

creasing the kinetic energy of the mass. (The useless work 
due to slip is inevitable in all paddle or screw propulsion, as 
well as a certain amount lost in machine-friction, not considered 
in the present problem.) During the " recover" the velocity 
decreases again to v Q . 

155. Examples. — 1. What work is done on a level track, in 
bringing up the velocity of a train weighing 200 tons, from 
zero to 30 miles per hour, if the total frictional resistance (at 
any velocity, say) is 10 lbs. per ton, and if the change of speed 
is accomplished in a length of 3000 feet ? 

{Foot-ton-second system.) 30 miles per hour == 44 ft. per 
sec. The mass 

= 200 -*- 32.2 = 6.2 ; 
.\ the change in kinetic energy, 

(=pfi,"_Pf X 2 ), 
= -|(6.2) X 44 2 = 6001.6 ft.-tons. 
11 



162 MECHANICS OF ENGINEERING. 

The work done in overcoming friction = Fs, i.e., 

= 200 X 10 X 3000 = 6,000,000 ft.-lbs. = 3000 ft.-tons ; 
.-. total work = 6001.6 -j- 3000 = 9001.6 ft.-tons. 

(If the track were an up-grade, 1 in 100 say, the item of 
200 X 30 = 6000 ft.-tons would be added.) 

Example 2. — Required the rate of work, or power, in Ex- 
ample 1. The power is variable, depending on the velocity of 
the train at any instant. Assume the motion to be uniformly 
accelerated, then the working force is constant ; call it P. 
The acceleration (§ 56) will be p=v*-r- 2s =1936-^6000= 0.322 
ft. per sq. sec; and since P — F = Mp, we have 

P = 1 ton -f- (200 -T- 32.2) X 0.322 = 3 tons, 

which is 6000 ~ 200 = 30 lbs. per ton of train, of which 20 is 
due to its- inertia, since when the speed becomes uniform the 
work of the engine is expended on friction alone. 

Hence when the velocity is 44 ft. per sec, the engine is 
working at the rate of Pv = 264,000 ft.-lbs. per sec, i.e., at the 
rate of 480 EL P.; 

At J of 3000 ft. from the start, at the rate of 240 H. P., half 
as much ; 

At a uniform speed of 30 miles an hour the power would be 
simply 1 X 44 = 44 ft. -tons per sec = 160 H. P. 

Example 3. — The resistance offered by still water to the 
passage of a certain steamer at 10 knots an hour is 15,000 lbs. 
What power must be developed by its engines, at this uniform 
speed, considering no loss in " slip" nor in friction of ma- 
chinery? Ans. 461 H. P. 

Example 4. — Same as 3, except that the speed is to be 15 
knots (i.e., nautical miles ; each = 6086 feet) an hour, assum- 
ing that the resistances are as the square of the speed (approxi- 
mately true). Am. 1556 IL P. 

Example 5. — Same as 3, except that 12$ of the power is ab- 
sorbed in the " slip" (i.e., in pushing aside and backwards the 
water acted on by the screw or paddle), and Sf in friction of 
machinery. Ans. 576 H. P. 

Example 6. — In Example 3, if the crank-shaft makes 60 



WORK, ENERGY, AND POWER. 163 

revolutions per minute, the crank-pin describing a circle of 18 
inches radius, required the average* value of the tangential 
component of the thrust (or pull) of the connecting-rod against 
the crank-pin. Ans. 26890 lbs. 

Example 7. — A solid sphere of cast-iron is rolling up an in- 
cline of 30°, and at a certain instant its centre has a velocity of 
36 inches per second. Neglecting friction of all kinds, how 
much further will the ball mount the incline (see § 143) ? 

Ans. 0.390 ft. 

Example 8. — In Fig. 163, with o = 4 f t. and a = 16 inches, 
it is found in one experiment that the friction which keeps the 
speed of the pulley at 120 revolutions per minute is balanced 
by a weight G — 160 lbs. Required the power thus measured. 

Ans. 14.6 H. P. 

Although in Examples 1 to 6 the steam cylinder is itself in 
motion, the work per stroke is still = mean effective steam- 
pressure on piston X length of stroke, for this is the final form 
to which the separate amounts of work done by, or upon, the 
two cylinder heads and the two sides of the piston will re- 
duce, when added algebraically. See § 154. 

* By " average value" is meant such a value, T m , as multiplied iuto the 
length of path described by the crank-pin per unit of time shall give the 
power exerted . 



It)4 



MECHANICS OF ENGINEERING. 



CHAPTER VII. 



FRICTION. 




156. Sliding Friction. — When the surfaces of contact of two 
bodies are perfectly smooth, the direction of the pressure or pair 
of forces between them is normal to these surfaces, i.e., to their 
tangent-plane ; but when they are rough, and 
moving one on the other, the forces or ac- 
tions between them incline away from the 
normal, each on the side opposite to the di- 
rection of the (relative) motion of the body 
on which it acts. Thus, Fig. 167, a block 
whose weight is 6r, is drawn on a rough 
horizontal table by a horizontal cord, the tension in which is 
P. On account of the roughness of one or both bodies the ac- 
tion of the table upon the block is a force P^ inclined to the 
normal (which is vertical in this case) at an angle = cp away 
from the direction of the relative velocity v. This angle cp is 
called the angle of friction, while the tangential component of 
P\ is called the friction — F. The normal component N, 
which in this case is equal and opposite to G the weight of the 
body, is called the normal pressure. 

Obviously F = i^tan cp, and denoting tan <p hjf we have 

F=fN. (1) 

f is called the coefficient of friction, and may also be defined 
as the ratio of the friction F to the normal pressure JV which 
produces it. 



FRICTION. 165 

In Fig. 167, if the motion is accelerated (ace. = jp), we have 
(eq. (IV.), § 55) P - F = Mp ; if uniform, P — F= 6 \ from 
which equations (see also (1))/ may be computed. In the 
latter case /. may be found to be different with different veloci- 
ties (the surfaces retaining the same character of course), and 
then a uniformly accelerated motion is impossible unless P 
— F were constant. 

As for the lower block or table, forces the equals and op- 
posites of J¥&ndF(or a single force equal and opposite to P^) 
are comprised in the sj'stem of forces acting upon it. 

As to whether F is a working force or a resistance, when 
either of the two bodies is considered free, depends on the cir- 
cumstances of its motion. For example,' in friction-gearing 
the tangential action between the two pulleys is a resistance 
for one, a working force for the other. 

If the force P, Fig. 167, is just sufficient to start the body, 
or is just on the point of starting it (this will be called impending 
motion), Fis called the friction of rest. If the body is at rest 
and P is not sufficient to start it, the tangential component will 
then be < the friction of rest, viz., just = P. As P increases, 
this component continually equals it in value, and P x acquires 
a direction more and more inclined from the normal, until the 
instant of impending motion, when the tangential component 
=f]¥= the friction of rest. When motion is once in prog- 
ress, the friction, called then the friction of motion, — fN, 
in which/ is not necessarily the same as in the friction of rest. 

157. Laws of Sliding Friction. — Experiment has demon- 
strated the following relations approximately, for two given 
rubbing surfaces : (See § 175.) 

(1) The coefficient,/ 1 , is independent of the normal pressure 
JST. 

(2) The coefficient,/, for friction of motion, is the same at 
all velocities. 

(3) The coefficient, f for friction of rest (i.e., impending 
motion) is usually greater than that for friction of motion 
(probably on account of adhesion). 



166 



MECHANICS OF ENGINEERING. 



(4) The coefficient,^ is independent of the extent of rub- 
bing surface. 

(5) The interposition of an unguent (such as oil, lard, tallow, 
etc.) diminishes the friction very considerably. 

158. Experiments on Sliding Friction. — These may be made 
with simple apparatus. If a block of weight = G, Fig. 168, 
be placed on an inclined plane of uniformly rough surface, 
and the latter be gradually more and more inclined from the 
horizontal until the block begins to move, the value of ft at 





Fig. 168. 

this instant = g>, and tan q> = f = coefficient of friction of 
rest. For from 2X =0 we have F, i.e., fN, = G sin ft; 
from 2Y= 0, iV= G cos ft ; whence tan ft = f, .'. ft must 

= <P- 

Suppose ft so great that the motion is accelerated, the body 

starting from rest at o, Fig. 169. It will be found that the 
distance x varies as the square of the time, hence (§ 56) the 
motion is uniformly accelerated (along the axis X). (Notice 
in the figure that G is no longer equal and opposite to P„ the 
resultant of iVand F, as in Fig. 168.) 

2 T = 0, which gives N — G cos ft = ; 

2X = Mp, which gives G sin ft—fN— (G -f- g)p ; 

while (from § 56) p=2x+- f. 

Hence, by elimination, x and the corresponding time t having 
been observed, we have for the coefficient of friction of motion 



f=t2LTift 



2x 



gtf cos ft' 



FRICTION. 167 

In view of (3), § 157, it is evident that if a value fi m has been 
found experimentally for j3 such that the block, once started by 
hand, preserves a uniform motion down the plane, then, since 
tan fi m =zf for friction of motion, J3 m may be less than the /3 
in Fig. 168, for friction of rest. 

159. Another apparatus consists of a horizontal plane, a pul- 
ley, cord, and two weights, as shown in Fig. 59. The masses 
of the cord and pulley being small and hence neglected, the 
analysis of the problem when G is so large as to cause an ac- 
celerated motion is the same as in that example [(2) in § 57], 
except in Fig. 60, where the frictional resistance yiV should be 
put in pointing toward the left. JV still = G iy and .*. 

S-fG 1 = (G^g)p; (1) 

while for the other free body in Fig. 61 we have, as before, 

e-s={e+Ap (2) 

From (1) and (2), S the cord-tension can be eliminated, and 
solving for p, writing it equal to 2s -=- f, s and t being the ob- 
served distance described (from rest) and corresponding time, 
we have finally for friction of motion 

_Q_ G+G, 2s 

t - a x ~ &—■& < 3 ) 

If G, Fig. 59, is made just sufficient to start the block, or 
sledge, G iy we have for the friction of rest 

/=! (*) 

160. Results of Experiments on Sliding Friction. — Professor 
Thurston in his article on Friction (which the student will do 
well to read) in Johnson's Cyclopaedia gives the following 
epitome of results from General Morin's experiments (made 
for the French Government in 1833) : 



168 



MECHANICS OF ENGINEERING. 
TABLE FOR FRICTION OF MOTION. 



No. 


Surfaces. 


Unguent. 


Angle <f>. 


/ = tan <f>. 


1 


Wood on wood. 


None. 


14° to 26£° 


0.25 to 0.50 


2 


"Wood on wood. 


Soap. 


2° to lli° 


0.04 to 0.20 


3 


Metal on wood. 


None. 


26£° to 3H° 


0.50 to 0.60 


4 


Metal on wood. 


Water. 


15° to 20° 


0.25 to 0.35 


5 


Metal on wood. 


Soap. 


nr 


0.20 


6 


Leather on metal. 


None. 


29 £° 


0.56 


7 


Leather on metal. 


Greased. 


13° 


0.23 


8 


Leather on metal. 


Water. 


20° 


0.36 


9 


Leather on metal. 


Oil. 


8F 


0.15 


10 


Smoothest and best 










lubricated surfaces. 




If to 2° 


0.03 to 0.036 









For friction of rest, about 4:0% may be added to the coeffi- 
cients in the above table. 

In dealing with the stone blocks of an arch-ring, cp is com- 
monly taken = 30°, i.e., f = tan 30° = 0.58 as a low safe 
value ; it is considered that if the direction of pressure between 
two stones makes an angle > 30 with the normal to the joint 
(see § 161) slipping may take place (the adhesion of cement 
being neglected). 

General Morin states that for a sledge on dry ground f = 
about 0.66. 

Weisbach gives for metal on metal, dry (K. R. brakes for 
example), f = from 0.15 to 0.24. Trautwine's Pocket-Book 
gives values of f for numerous cases of friction. 

161. Cone of Friction.— Fig. 170. Let A and B be two 

rough blocks, of which B is immovable, and P the resultant 
\0/ r °f a H the forces acting on A, except the pres- 
sure from B. B can furnish any required 
normal pressure iV to balance P cos /?, but 
the limit of its tangential resistance is fN. 
So long then as /? is < cp the angle of fric- 
tion, or in other words, so long as the line of 
fig. i?o. action of P is within the u cone of friction" 

generated by revolving OC about ON, the block A will not 




FRICTION. 169 

slip on B, and the tangential resistance of B is simply P sin 
p ; but if fi is > <p, this tangential resistance being only fN 
and < P sin /?, 4 will begin to slip, with an acceleration. 

162. Problems in Sliding Friction. — In the following prob- 
lems/* is supposed known at points where rubbing occurs, or 
is impending. As to the pressure N to which the friction is 
due, it is generally to be considered unknown until determined 
by the conditions of the problem. Sometimes it may be an 
advantage to deal with the single unknown force P (resultant 
of iTandyiV) acting in a line making the known angle cp with 
the normal (on the side away from the motion). 

Problem 1. — Eequired the value of the weight P, Fig. 171, 
the slightest addition to which will cause motion of the hori- 
zontal rod OB, resting on rough planes at 45°. The weight 
G of the rod may be applied at the /^y 

middle. Consider the rod free ; at 
each point of contact there is an un- 
known N and a friction due to it 
fN\ the tension in the cord will be 
= P, since there is no acceleration 
and no friction at pulley. Notice fig. m. 

the direction of the frictions, both opposing the impending 
motion. [The student should not rush to the conclusion that 
JV and JST X are equal, and are the same as would be produced by 
the components of G if the latter were transferred to A and 
resolved along AO and AB ; but should await the legitimate 
results deduced by algebra, from the equations of condition 
for the equilibrium of a system of forces in a plane. Few 
problems in Mechanics are so simple as to admit of an imme- 
diate mental solution on inspection ; and guess-work should be 
carefully avoided.] 

Taking an origin and two axes as in figure, we have (eqs. 
(2), § 36), denoting the sine of 45° by m, 

2X fN^+mG-N -P = 0;. . (1) 

2Y JY^fJST- mG = 0;. . (2) 

2(Pa) fWa + Wa- Gb = 0. . . (3) 



170 



MECHANICS OF ENGINEERING. 



The three unknowns P, JV, and J¥ x can now be found. 
Divide (3) by a, remembering that b : a = m, and solve for 
JV; substitute it in (2) and JV r also becomes known ; while P 
is then found from (1) and is 



P = 



2infG _ /V2 
l+f-'l+f 



. G. 




Problem 2. — Fig. 172. A rod, centre of gravity at middle, 
leans against a rough wall, and rests on an equally rough floor; 
how small may the angle a become before it 
slips ? Let a = the half-length. The figure 
shows the rod free, and following the sugges- 
tion of § 162, a single unknown force P x 
making a known angle cp (whose tan =,/*)• 
with the normal BE, is put in at B, leaning 
away from the direction of the impending 
motion, instead of an JV and fN\ similarly 
P 2 acts at 0. The present system consisting 
of but three forces, the most direct method of finding or, with- 
out introducing the other two unknowns P x and P 7 at all, is 
to use the principle that if three forces balance, their lines 
of action must intersect in a point. That is, P^ must inter- 
sect the vertical containing G, the weight, in the same point 
as P x , viz., A. 

Now PA, and also BO, — a cos a, 

.'. ED = a cos a cot cp and AB = a cos a tan cp. 

But BE, which = 2a sin or, = BE— AB '; 

.*. 2a sin a = a cos a [cot cp — tan cp\ . . (1) 

Dividing by cos a, and noting that tan cp =\f= 1 -r- cot <p, 
we obtain for the required value of a 



tan a = «■ . 



i i-r 



f 



and finally, tan a = cot 2<p, 



after some trigonometrical reduction. That is, a is the com- 
plement of double the angle of friction. 



FEICTION. 



171 



acting 



Problem 3. — Fig. 173. Given the resistance Q 
parallel to the fixed guide C, the angle a, and the (equal) co- 
efficients of friction at the rubbing surfaces, required the 





Fig. 173. 

amount of the horizontal force P, at the head of the block A 
(or wedge), to overcome Q and the frictions, D is fixed, and 
ah is perpendicular to cd. Here we have four unknowns, viz., 
P, and the three pressures i\T, JV V and JV 2 , between the blocks. 
Consider A and B as free bodies, separately (see Fig. 174), re- 
membering Newton's law of action and reaction. The full 
values (e.g.,/!^) of the frictions are put in, since we suppose 
a slow uniform motion taking place. 
For A, 2X= and 2 Y = give 



i\T — iV^cos a -\-fJVsin a — P sin a = ; . 
fN x + N sin a -j-yiTcos a — P cos a = 0. . 
For^, ^Xand .2 T give 
0-^+/-^ 3 = O;....(S) and ^ 2 -/AT, = 0.' 
Solve (4) for iT a and substitute in (3), whence 



(i) 

(2) 
•(4) 

(5) 

Solve (2) for iV, substitute the result in (1), as also the value 
of N x from (5), and the resulting equation contains but one un- 
known, P. Solving for P, putting for brevity 



cos a 



we have 



or 



ycos a -f- sin a = m and 

(m+fn)Q 



■fsin a 



n< 



P = 



(n . cos a -\- m . sin <*)(1 —f*)' 
P=Q(m+fn) -HI-/')- . 



(6) 

(7) 



172 MECHANICS OF ENGINEERING. 

Numerical Example of Problem 3. — If Q = 120 lbs., f 
= 0.20 {an abstract number, and .'. the same in any system of 
units), while a = 14°, whose sine = 0.240 and cosine -= .970, 
then 

m = 0.2x.97 + 0.24 = 0.43 and n = .97 -.2X.24 = 0.92, 

whence P — QMQ = -76.8 lbs. 

While the wedge moves 2 inches P does the work (or exerts 
an energy) of 2 X 76.80 = 153.6 in.-lbs. = 12.8 ft.-lbs. 

For a distance of 2 inches described by the wedge horizon- 
tally, the block B (and .*. the resistance Q) has been moved 
through a distance = 2 X sin 14° = 0.48 in. along the guide 
G, and hence the work of 120 X 0.48 = 57.6 in.-lbs. has been 
done upon Q. Therefore for the supposed portion of the 
motion 153.6 — 57.6 = 96.0 in.-lbs. of work has been lost in 
friction (converted into heat). 

It is noticeable in eq. (6), that if f should = 1.00, P = oc ; 
and that if a == 90°, P = Q, and there is no friction (the 
weights of the blocks have been neglected). 

Problem 4. Numerical. — With what minimum pressure 

P should the pulley A be held against B. which it drives by 

i p| ,^ p. A " frictional gearing," to transmit 2 H.P.; 

•"7 P *;'- js\ ) if or = 45°, f for impending (relative) 

a[vj _.,--'n LJ motion, i.e., for impending slipping = 

fig. 175. 0.40, and the velocity of the pnlley-rim 

is 9 ft. per second ? 

The limit-value of the tangential " grip" 

T = 2fN= 2 X 0.40 X P sin 45°, 
2 H. P. = 2 X 550 = 1100 ft.-lbs. per second. 

Putting T X 9 ft. = 1100, we have 

2 X 0.40 x4xPx9 = 1100 ; :,P= 215 lbs. 

Problem 6. — A block of weight G lies on a rough plane, 
inclined an angle /? from the horizontal ; find the pull P, mak- 
ing an angle a with the first plane, which will maintain a uni- 
form motion up the plane. 



FRICTION. 173 

Problem 7. — Same as 6, except that the pull P is to permit 
a uniform motion down the plane. 

Problem 8. — The thrust of a screw-propeller is 15 tons. 
The ring against which it is exerted has a mean radius of 8 
inches, the shaft makes one revolution per second, andy= 0.06. 
Required the H. P. lost in friction from this cause. 

Ana. 13.7 H. P. 

163. The Bent-Lever with Friction. Worn Bearing. — Fig. 
176. Neglect the weight of the lever, and suppose the plumb- 
er-block so worn that there is d*- H E 

contact along one element only of 
the shaft. Given the amount and 
line of action of the resistance P, 
and the line of action of P, re- 
quired the amount of the latter for 
impending slipping in the direction 
of the dotted arrow. As P grad- 
ually increases, the shaft of the 
lever (or gear-wheel) rolls on its 5^T 

bearing until the line of contact has reached some position A 9 
when rolling ceases and slipping begins. To find A, and the 
value of P, note that the total action of the bearing upon the 
lever is some force P„ applied at A and making a known 
angle cp (f = tan cp) with the normal A G. P 1 must be equal 
and opposite to the resultant of the known R and the unknown 
jP, and hence graphically (a graphic is much simpler here than 
an analytical solution) if we describe about (7 a circle of radius 
= r sin cp, r being the radius of shaft (or gudgeon), and draw 
a tangent to it from P, we determine PA as the line of action 
of P v If PG is made = P, to scale, and GF drawn parallel 
to P . . . P, P is determined, being = PF, while P 1 ~ PF. 

If the known force P is capable of acting as a working force, 
by drawing the other tangent PB from P to the " friction- 
circle," we have P = PH, and P x = PK, for impending 
rotation in an opposite direction. 

If P and P are the tooth-pressures upon two spur-wheels, 
keyed upon the same shaft and nearly in the same plane, the 




174 



MECHANICS OF ENGINEERING. 



[P 1 sin cp]Z7rr. 



same constructions hold good, and for a continuous uniform 
motion, since the friction = P t sin q>, 
the work lost in friction 
per revolution, 

It is to be remarked, that without friction P x would pass 
through G, and that the moments of P and P would balance 
about G (for rest or uniform rotation) ; whereas with friction 
they balance about the proper tangent-point of the friction- 
circle. 

Another way of stating this is as follows: So long as the 
resultant of P and P falls within the " dead-angle" PDA, 
motion is impossible in either direction. 

If the weight of the lever is considered, the resultant of it 
and the force P can be . substituted for the latter in the fore- 
going. 

164. Bent-Lever with Friction. Triangular Bearing. — Like 
the preceding, the gudgeon is much exaggerated in the figure 

(177). For impending rotation in 
direction of the force P, the total 
actions at A 1 and A 2 must lie in 
known directions, making angles = <p 
with the respective normals, and in- 
clined away from the slipping. Join 
the intersections D and L. Since 
the resultant of P and P at D must 
act along DL to balance that of P x 
and jP 2 , having given one force, say 
P, we easily find P = DE, while 
P 1 and P 2 = ZJI^and LN respectively, LG having been made 
= PP, and the parallelogram completed. 

(If the direction of impending rotation is reversed, the change 
in the construction is obvious.) If P 2 = 0, the case reduces 
to that in Fig. 176 ; if the construction gives P % negative, the 
supposed contact at A a is not realized, and the angle A i GA 1 
should be increased, or shifted, until jP 9 is positive. 

As before, P and P may be the tooth-pressures on two 




Fig. 177. 



FRICTION. 175 

spur-wheels nearly in the same plane and on the same shaft ; 
if so, then, for a uniform rotation, 
Work lost in fric. per re vol. = [P 1 sin cp + P t sin qPfinr. 

165. Axle*Friction. — The two foregoing articles are intro- 
ductory to the subject of axle -friction. When the bearing is 
new, or nearly so, the elements of the axle which are in contact 
with the bearing are infinite in number, thus giving an infinite 
number of unknown forces similar to P x and P % of the last 
paragraph, each making an angle cp with its normal. Refined 
theories as to the law of distribution of these pressures are of 
little use, considering the uncertainties as to the value of 
y ( = tan cp) ; hence for practical purposes axle-friction may be 
written 

F=fB, 

in which f is a coefficient of axle-friction derivable from 
experiments with axles, and P the resultant pressure on the 
bearing. In some cases P may be partly due to the tightness 
of the bolts with which the cap of the bearing is fastened. 

As before, the work lost in overcoming axle-friction jper 
revolution is =.fP%7tr, in which r is the. radius of the axle. 
f\ like f is an abstract number. As in Fig. 176, a u friction- 
circle," of radius =fr, may be considered as subtending the 
" dead-angle." 

166. Experiments with Axle-Friction. — Prominent among 
recent experiments have been those 
of Professor Thurston (1872-73), 
who invented a special instrument 
for that purpose, shown (in princi- 
ple only) in Fig. 178. By means of 
an internal spring, the amount of 
whose compression is read on a scale, 
a weighted bar or pendulum is caused 
to exert pressure on a projecting axle 
from which it is suspended. The 
axle is made to rotate at any desired 
velocity by some source of power, the axle-friction causing 




176 MECHANICS OF ENGINEERING. 

the pendulum to remain at rest at some angle of deviation! 
from the vertical. The figure shows the pendulum free, the 
action of gravity upon it being G, that of the axle consisting 
of the two pressures, each = i?, and of the two frictions (each 
being F =f R\ due to them. Taking moments about C, we 
have for equilibrium 

2fRr = Gb, 

in which all the quantities except f are known or observed. 
The temperature of the bearing is also noted, with reference 
to its effect on the lubricant employed. Thus the instrument 
covers a wide range of relations. 

General Morin's experiments as interpreted by Weisbach 
give the following practical results : 

■\ r 0.054 for well-sustained 

For iron axles, in iron or I lubrication; 

brass bearings ^ j 0.07 to .08 for ordinary 

I lubrication. 



By " pressure per square inch on the bearing" is commonly 
meant the quotient' of the total pressure in lbs. by the area in 
square inches obtained by multiplying the width of the axle by 
the length of bearing (this length is quite commonly four times 
the diameter) ; call it j?, and the velocity of rubbing in feet per 
minute, v. Then, according to Rankine, to prevent overheat- 
ing, we should have 

p(v + 20) < 44800 . . . (not homog.). 

Still, in marine-engine bearings pv alone oftemreaches 60,000, 
as also in some locomotives (Cotterill). Good practice keeps 
p within the limit of 800 (lbs. per sq. in.) for other metals 
than steel (Thurston), for which 1200 is sometimes allowed. 

With v ■=■ 200 (feet per min.) Professor Thurston found that 
for ordinary lubricants p should not exceed values ranging 
from 30 to 75 (lbs. per sq. in.). 

The product pv is obviously proportional to the power ex- 
pended in wearing the rubbing surfaces, per unit of area. 



FRICTION. 



177 




167. Friction-Wheels. — A single example of their use will 
be given, with some approximations to avoid complexity. Fig. 
179. G is the weight of a heavy wheel, P x is a known vertical 
resistance (tooth-pressure), and P an 
unknown vertical working force, 
whose value is to be determined to 
maintain a uniform rotation. The 
utility of the friction-wheels is also 
to be shown. The resultant of P x , 
G, and P is a vertical force i?, pass- 
ing nearly through the centre C of 
the main axle which rolls on the four 
friction-wheels, i?, resolved along 
CA and CP, produces (nearly) equal 
pressures, each being N = P -j- 2 cos ar, at the two axles of 
the friction- wheels, which rub against their fixed plumber- 
blocks. P = P -\- P x -f- G x , and .*. contains the unknown P, 
but approximately = G -\- %P X , i.e., is nearly the same (in this 
case) whether friction-wheels are employed or not. 

When G makes one revolution, the friction f'N at each axle 
C x is overcome through a distance = (r x : a x ) 27tr, and 

"Work lost per revol. j 
with 
friction-wheels, ] 

Whereas, if C revolved in a fixed bearing, 

Work lost per revol. 

without 

friction-wheels, 



Of 'P 

2 f'N ^nr = -'-- 
J a x a x cos a 



fP27rr. 



fPZTtr. 



Apparently, then, there is a saving of work in the ratio r x : 
a x cos a, but strictly the 7?is not quite the same in the two cases ; 
for with friction-wheels the force P is less than without, and R 
depends on P as well as on the known G and P x . By dimin- 
ishing the ratio r x : a x , and the angle <*, the saving is increased. 
If a were so large that cos a < r x : a x , there would be no saving, 
but the reverse. 

As to the value of P to maintain uniform rotation, We have 
12 



178 MECHANICS OF ENGINEERING. 

for equilibrium of moments about C, with friction-wheels (con. 
sidering the large wheel and axle free), 

JPb = P l b l + 2Tr, (1) 

in which T is the tangential action, or " grip," between one 
pair of friction -wheels and the axle C which rolls upon them. 
T would noc equal fN unless slipping took place or were im- 
pending at E, but is known by considering a pair of friction- 
wheels free, when 2 {Pa) about 1 gives 

R r 
Ta x =fJSTr 1 =f~ . -A-, 
1 J y J 2 cos 0? 

which in (1) gives finally 

h r r 
P = T l P 1 -\ -^—fP, (2) 

b ' ' a x cos ay b v ' 

Without friction-wheels, we would have 

P = %JP,+fl% (3) 

The last term in (2) is seen to be less than that in (3) (unless 
a is too large), in the same ratio as already found for the saving 
of work, supposing the P's equal. 

If P x were on the same side of C as P, it would be of an 
opposite direction, and the pressure P would be diminished. 
Again, if P were horizontal, P would not be vertical, and the 
friction- wheel axles would not bear equal pressures. Since P 
depends on P l5 (7, and the frictions, while the friction depends 
on P, and P on P 1? G, and P, an exact analysis is quite 
complex, and is not warranted by its practical utility. 

Example. — If an empty vertical water-wheel weighs 25,000 
lbs., required the force P to be applied at its circumference to 
maintain a uniform motion, with a = 15 ft., and r — 5 inches. 
Here P x — 0, and P = G (nearly ; neglecting the influence of 
P on P), i.e., P = 25,000 lbs. 

First, without friction-wheels (adopting the foot-pound-sec- 
ond system of units), with t / y = .07 (abstract number). From 
eq. (3) we have 

P = + 0.07 X 25,000 X (A + 15 ) = 48 - 6 lbs - 



FRICTION. 179 

The work lost in friction per revolution is 
fPZnr = 0.07 X 25,000 X 2 X 3.14 X A = 4580 ft - lbs - 
Secondly, with friction-wheels, in which r x : a x = j> and 
cos a = 0.80 (i.e., or = 36°). From eq. (2) 

P = + | . J^- X 48.6 = only 12.15 lbs., 
while the work lost per revolution 

= i • ¥ X 4580 = 1145ft.-lbs. 
Of course with friction- wheels the wheel is not so steady as 
without. 

In this example the force P has been simply enough to 
overcome friction. In case the wheel is in actual use, P is the 
weight of water actually in the buckets at any instant, and does 
the work of overcoming P xt the resistance of the mill machinery, 
and also the friction. By placing P x pointing upward on the 
same side of C as P, and making b x nearly = o, P will = G 
nearly, just as when the wheel is running empty; and the 
foregoing numerical results will still hold good for practical 
purposes. 

168. Friction of Pivots. — In the case of a vertical shaft or 
axle, and sometimes in other cases, the extremity requires sup- 
port against a thrust along the axis of the axle or pivot. If 
the end of the pivot is flat and also the surface 
against which it rubs, we may consider the 
pressure, and therefore the friction, as uniform 
over the surface. With a flat circular pivot, 
then, Fig. 180, the frictions on a small sector 
of the circle form a system of parallel forces 
whose resultant is equal to their sum, and is FlG * m 
applied a distance of %r from the centre. Hence the sum of 
the moments of all the frictions about the centre =fP%r, in 
which P is the axial pressure. Therefore a force P necessary 
to overcome the friction with uniform rotation must have a 
moment 

Pa=fR\r, 




180 



MECHANICS OF ENGINEERING. 



and the work lost in friction per revolution is 
=/£2* . fr = tnfRr. 



(1) 



As the pivot and step become worn, the resultant frictions 
in the small sectors probably approach the centre ; for the 
greatest wear occurs first near the outer edge, since there the 
product^ is greatest (see § 166). Hence for %r we may more 
reasonably put -Jr. 

Example. — A vertical flat-ended pivot presses its step with 
a force of 12 tons, is 6 inches in diameter, and makes 40 revolu- 
tions per minute. Eequired the H. P. absorbed by the friction. 
Supposing the pivot and step new, and f for good lubrication 
= 0.07, we have, from eq. (1) {foot-lb. -second), 

Work lost per revolution 



.07 X 24,000 X 6.28 X * . i = 1758.4 ft.-lbs., 



and 



work per second 
= 1758.4 X U 



= 1172.2 ft.-lbs., 



If ordi- 



which -r- 550 gives 2.13 H. P. absorbed in friction. 

nary axle-friction also occurs its effect must be added. 

If the flat-ended pivot is hollow, with radii r 1 and r„ we may 

put i(r x -f- r z ) instead of the %r of the preceding. 

It is obvious that the smaller the lever-arm given to the 

resultant friction in each sector of the rubbing surface the 

smaller the power lost in friction. Hence pivots'should be 

made as small as possible, consistently with strength. 

For a conical pivot and step, Fig. 181, the resultant friction 
in each sector of the conical bearing surface has 
a lever-arm =s \r x about the axis A, and a value 
> than for a flat-ended pivot ; for, on account 
of the wedge-like action of the bodies, the 
pressure causing friction is greater. The sum of 
the moments of these resultant frictions about 
A is the same as if only two elements of the 

cone received pressure (each = JV = \R -r- sin a). Hence the 




FKICTION. 181 

moment of friction of the pivot, i.e., the moment of the force 
necessary to maintain uniform rotation, is 

Pa =f2JV~r 1 =fJ^- %r» 
J 3 x • / sina3 " 

and work lost per revolution == Fr n f~ ?V 

r 3 J sm a x 

By making r x small enough, these values may be made less 
than those for a flat-ended pivot of the same diameter = 2r. 

In Schiele's " anti-friction" pivots the outline is designed 
according to the following theory for securing uniform vertical 
wear. Let p = the pressure per p"~^ ~~~| — - — -^ 

horizontal unit of area (i.e., p jA \ p 

= R -f- horizontal projection of .^ — 1 f]^illlllll& 

the actual rubbing surface) : / ifB£m > i V^^liP 5 ' 

this is assumed constant. Let ^Jwfor^, jltt.-V 

the unit of area be small, for j| 'ic K 

algebraic simplicity. The fric- fkj. i82. 

tion on the rubbing surface, whose horizontal projection = unity, 
is =fJ¥=f(j> h- sin a) (see Fig. 182; the horizontal com- 
ponent of jp is annulled by a corresponding one opposite). The 
work per revolution in producing wear on this area = fN^ny. 
But the vertical depth of wear per revolution is to be the same 
at all parts of the surface ; and this implies that the same 
volume of material is worn away under each horizontal unit of 

area. Hence fN^ny, i.e.,f-r^ — 27ry, is to be constant for all 

values of y ; and since fp and 2?r are constant, we must have, 
as the law of the curve, 

, i.e., the tangent BC = the same at all points. 



sm a 



This curve is called the " tractrixP Schiele's pivots give a 
very uniform wear at high speeds. The smoothness of wear 
prevents leakage in the case of cocks and faucets. 

169. Normal Pressure of Belting. — When a perfectly flexible 
cord, or belt, is stretched over a smooth cylinder, both at rest, 



182 



MECHANICS OF ENGINEERING. 



X- 




the action between them is normal at every point. As to its 
j&- \_\ t t s am<)lirit ? Pi P er linear unit of arc, the f ol- 
'^ = I r * lowing will determine. Consider a semi- 
circle of the cord free, neglecting its weight. 
Fig. 183. The forces holding it in equilib- 
rium are the tensions at the two ends (these 
are equal, manifestly, the cylinder being 
smooth ; for they are the only two forces 
^/nr 3 *" having moments about (7, and each has the 
fig. 183. same lever-arm), and the normal pressures, 

which are infinite in number, but have an intensity, p, per 
linear unit, which must be constant along the curve since 8 is 
the same at all points. The normal pressure on a single ele- 
ment, ds, of the cord is = pds, and its X component = 
pds cos 6 z=z prdd cos 0. Hence 2X = gives 

rp f cos ddO — 28 = 0, i.e., rp\ sin = 28; 

*J~ in L- in 



TPl 1 — (— !)] = 2 # or p = 



8 



(1) 



170. Belt on Rough Cylinder. Impending Slipping. — If fric- 
tion is possible between the two bodies, the tension may vary 
along the arc of contact, so that j> also varies, and consequently 




Fig. 184. 




the friction on an element ds being =f<pds =f(S-±- r)ds, also 
varies- If slipping is impending, the law of variation of the 
tension 8 may be found, as follows : Fig. 184, in which the 



FKICTION. 



183 



impending slipping is toward the left, shows the cord free. 
For any element, ds, of the cord, we have, putting 2 (moms, 
about <9) = (Fig. 185), 

(8+ dS)r = 8r + dFr ; i.e., dF= dS, 

or (see above) dS =f(S -r- r)ds. 

But ds = rdd ; hence, after transforming, 



AM dS 

fdd = -£-. 



(1) 



In (1) the two variables 6 and 8 are separated ; (1) is there- 
fore ready for integration. 

>=log e ^-log e ^ = log 6 [|]. (2) 

Or, by inversion, 8 Q ef* — 8 n , (3) 

e, denoting the Naperian base, = 2.71828 +; a of course is in 
^"-measure. 

Since 8 n evidently increases very rapidly as a becomes 
larger, 8 a remaining the same, we have the explanation of the 
well-known fact that a comparatively small tension, 8 , exerted 
by a man, is able to prevent the slipping of a rope around a 
pile-head, when the further end is under the great tension 8 n 
due to the stopping of a moving steamer. For example, with 
f — ^, we have (Weisbach) 

f or a = \ turn, or a = \n, 8 n — 1.69# ; 

= \ turn, or a = tt, 8 n = 2.85# ; 

= 1 turn, or a = 2tt, 8 n = 8.12S ; 

= 2 turns, or a = 4tt, 8 n = 65.94# ; 

= 4 turns, or a = 8tt, S n = 4348.56# . 

If slipping actually occurs, we must use a value of f for fric- 
tion of motion. 

Example. — A leather belt drives an iron pulley, covering 
one half the circumference. What is the limiting value of the 



184 



MECHANICS OF ENGINEERING. 



ratio of S n (tension on driving-side) to S (tension on follow- 
ing side) if the belt is not to slip, taking the low value of 
f = 0.25 for leather on iron ? 

We have given fa = 0.25 X n = .7854, which by eq. (2) is 
the Naperian log. of (S n : S a ) when slipping occurs. Hence the 
common log. of (S n : S Q ) = 0.7854 X 0.43429 = 0.34109 ; i.e., 
if 

(S n : S ) = 2.193, say 2.2, 
the belt will slip (for/= 0.25). 

(0.43429 is the modulus of the common system of loga* 
rithms, and = 1 : 2.30258. See example in §48.) 

At very high speeds the relation p = 3 ~ r (in § 169) is not 
strictly true, since the tensions at the two ends of an element 
ds are partly employed in furnishing the necessary deviating 
force to keep the element of the cord in its circular path, the 
remainder producing normal pressure. 

171. Transmission of Power by Belting or Wire Rope. — In the 

simple design in Fig. 186, it is required to find the motive 
weight G, necessary to overcome the given resistance R at a 



DRIVING SIDE 




Fig. 186. 



uniform velocity = v x \ also the proper stationary tension 
weight G to prevent slipping of the belt on its pulleys, and 
the amount of power, Z, transmitted. 
In other words, 



Given : 



( R, a, r, a„ rj a == n for both pulleys, ) 
( Vj] andy for both pulleys ; i 



-r> . i . j L ; G, to furnish L ; G for no slip ; v the velocity 
' I of G ; v' that of belt ; and the tensions in belt. 



FRICTION. 



185 



Neglecting axle-friction and the rigidity of the belting, the 
power transmitted is that required to overcome R through a 
distance = v x every second, i.e., 

Z = Rv, (1) 

Since (if the belts do not slip) 



a : r::v 



we have 



0, 



and. 2 : r 1 : : v f : v 1 



and v = v.. 

a r 1 1 



(2) 



Neglecting the mass of the belt, and assuming that each pul- 
ley revolves on a gravity-axis, we obtain the following, by con- 
sidering the free bodies in Fig. 187 : 




tfA fxee) 



(B free) 
Fig. 187. 



(B and truck free) 



2 (moms.) = in A free gives Rr x = (S n — S )a t ; 
2 (moms.) = in B free gives Gr = (S n — S )a ; 



(3) 



whence we readily find 



r a. 



Evidently R and G are inversely proportional to their velo- 
cities v 1 and v ; see (2). This ought to be true, since in Fig. 
186 G is the only working-force, R the only resistance, and 
the motions are uniform ; hence (from eq. (XYL), § 142) 

Gv - Rv, = 0. 

2X = 0, for B and truck free, gives 

#o = &n + #o> (5) 

while, for impending slip, 

Sn = S,ef* (6) 



186 MECHANICS OF ENGINEERING. 

By elimination between (4), (5), and (6), we obtain 

r_ ef + 1 _ L ef + 1 

~ a ' e?« - 1 ~~ v' ' ~e?" - V ' ' ' {i) 

L e* 
and 8 n = g, . efv ;_ 1 (8) 

Hence #„ and S n vary directly as the power transmitted and 
inversely as the velocity of the belt. For safety G should be 
made > the above value in (7) ; corresponding values of the 
two tensions may then be found from (5), and from the rela- 
tion (see § 150) 

(S n -S )v' = Z. ...... (6a) 

These new values of the tensions will be found to satisfy the 
condition of no slip, viz., 

(£„:#„)< ^(§170). 

For leather on iron, ef" = 2.2 (see example in § 170), as a 
low value. The belt should be made strong enough to with- 
stand S n safely. 

As the belt is more tightly stretched, and hence elongated, 
on the driving than on the following side, it " creeps" back- 
ward on the driving and forward on the driven pulley, so that 
the former moves slightly faster than the latter. The loss of 
work due to this cause does not exceed 2 percent with ordinary 
belting (Cotterill). 

In the foregoing it is evident that the sum of the tensions in 
the two sides = G , i.e., is the same, whether the power is 
being transmitted or not ; and this is found to be true, both in 
theory and by experiment, when a tension-weight is not used, 
viz., when an initial tension S is produced in the whole belt 
before transmitting the power, then after turning on the latter 
the sum of the two tensions (driving and following) always 
= 2$, since one side elongates as much as the other contracts ; 
it being understood that the pulley-axles preserve a constant 
distance apart. 

172. Rolling Friction. — The few experiments which have 
been made to determine the resistance offered by a level road- 



FRICTION. 



187 



way to the uniform motion of a roller or wheel rolling upon it 
corroborate approximately the following theory. The word 
friction is hardly appropriate in this connection (except when 
the roadway is perfectly elastic, as will be seen), but is sanctioned 
by usage. 

First, let the roadway or track be compressible, but inelastic, 
G the weight of the roller and its load, and P the horizontal 
force necessary to preserve a uniform motion 
(both of translation and rotation). The track 
(or roller itself) being compressed just in 
front, and not reacting symmetrically from 
behind, its resultant pressure against the 
roller is not at vertically under the centre, 
but some small distance, OD = b, in front. (The successive 
crushing of small projecting particles has the same effect.) 
Since for this case of motion the forces have the same relations 
as if balanced (see § 124), we may put 2 moms, about D = 0, 




Fig. 188. 



Pr = Gb ; or, 



P = -G. 
r 



(i) 



Coulomb found for 

Boilers of lignum-vitse on an oak track, b — 0.0189 inches; 
Rollers of elm on an oak track, b = 0.0320 inches. 

Weisbach's experiments give, for cast-iron wheels 20 inches in 
diameter on cast-iron rails, 

b = 0.0183 inches ; 

b = 0.0193 inches. 

>.4 inches in diameter, 

b = 0.0196 to 

0.0216 inches. 

According to the foregoing theory, P, the " rolling friction" 
(see eq. (1)), is directly proportional to G, and inversely to the 
radius, if b is constant. The experiments of General Morin and 
others confirm this, while those of Dupuit, Poiree. and Sauvage 
indicate it to be proportional directly to G, and inversely to the 
square root of the radius. 



and Rittinger, for the same, 
Pambour gives, for iron railroad wheels 



188 MECHANICS OF ENGINEEKING. 

Although b is a distance to be expressed in linear units, and 
not an abstract number like the f and f for sliding and axle- 
friction, it is sometimes called a " coefficient of rolling fric- 
tion." In eq. (1), b and r should be expressed in the same 
unit. 

Of course if P is applied at the top of the roller its lever- 
arm about D is 2r instead of r, with a corresponding change 
in eq. (1). 

With ordinary railroad cars the resistance due to axle and 
rolling frictions combined is about 8 lbs. per ton of weight on 
a level track. For wagons on macadamized roads b = \ inch, 
but on soft ground from 2 to 3 inches. 

Secondly, when the roadway is perfectly elastic. This is 
chiefly of theoretic interest, since at first sight no force would 
be considered necessary to maintain a uniform rolling motion. 
But, as the material of the roadway is compressed under the 
roller its surface is first elongated and then recovers its former 
state ; hence some rubbing and consequent sliding friction must 



Fig. 189. 



occur. Fig. 189 gives an exaggerated view of the circum- 
stances, P being the horizontal force applied at the centre 
necessary to maintain a uniform motion. The roadway (rub- 
ber, for instance) is heaped up both in front and behind the 
roller, being the point of greatest pressure and elongation 
of the surface. The forces acting are G, P, the normal 
pressures, and the frictions due to them, and must form a 
balanced system. Hence, since G and P, and also the normal 
pressures, pass through (7, the resultant of the frictions must 
also pass through C\ therefore the frictions, or tangential 
actions, on the roller must be some forward and some backward 



FRICTION. 



189 



(and not all in one direction, as seems to be asserted on p. 260 
of Cotterill's Applied Mechanics, where Professor Keynolds' 
explanation is cited). The resultant action of the roadway 
upon the roller acts, then, through some point J), a distance 
OD = b ahead of 0, and in the direction DO, and we have as 
before, with D as a centre of moments, 



Pr = Gb, or 



P = -G. 

r 



ti rolling friction is encountered above as 
well as below the rollers, Fig. 190, the 
student may easily prove, by considering 
three separate free bodies, that for uniform 
motion 



p = *-±4*, 



2r 




Fig. 190. 



where b and b 1 are the respective " coefficients of rolling fric- 
tion" for the upper and lower contacts. 

Example 1. — If it is found that a train of cars will move 
uniformly down an incline of 1 in 200, gravity being the only 
working force, and friction (both rolling and axle) the only 
resistance, required the coefficient, jT, of axle-friction, the 
diameter of all the wheels being 2r = 30 inches, that of the 
journals 2a = 3 inches, taking b = 0.02 inch for the rolling 
friction. Let us use equation (XYI.) (§ 142), noting that while 
the train moves a distance s measured on the incline, its weight 

G does the work G ~~~ s, the rolling friction — G (at* the axles) 

has been overcome through the distance s, and the axle-friction 

(total) through the (relative) distance — s in the journal boxes ; 

whence, the change in kinetic energy being zero, 

-L Gs- h Gs--fGs=0. 
200 r r J 

Gs cancels out, the ratios b : r and a : r are = y^-jj- and -^ 
respectively (being ratios or abstract numbers they have the 



* That is, the ideal resistance, at centre of axles and || to the incline, equiv- 
alent to actual rolling resistance. 



190 MECHANICS OF ENGINEERING. 

same numerical values, whether the inch or foot is used), and 
solving, we have 

/ = 0.05 — 0.0133 = 0.036. 
Example 2. — How many pounds of tractive effort per ton 
of load would the train in Example 1 require for uniform mor 
tion on a level track % Ans. 10 lbs. 

173. Railroad Brakes. — During the uniform motion of a 
railroad car the tangential action between the track and each 
wheel is small. Thus, in Example 1, just cited, if ten cars of 
eight wheels each make up the train, each car weighing 20 tons, 
the backward tangential action of the rails upon each wheel is 
only 25 lbs. When the brakes are applied to stop the train 
this action is much increased, and is the only agency by which 
the rails can retard the train, directly or indirectly : directly, 
when the pressure of the brakes is so great as to prevent the 
wheels from turning, thereby causing them to " skid " (i.e., 
slide) on the rails ; indirectly, when the brake-pressure is of 
such a value as still to permit perfect rolling of the wheel, in 
which case the rubbing (and heating) occurs between the brake 
and wheel, and the tangential action of the rail has a value 
equal to or less than the friction of rest. In the first case, 
then (skidding), the retarding influence of the rails is the, fric- 
tion of motion between rail and wheel ; in the second, a force 
which may be made as great as the friction of rest between rail 
and wheel. Hence, aside from the fact that skidding produces 
objectionable flat places on the wheel- tread, the brakes are 
more effective if so applied that skidding is impending, but 
not actually produced ; for the friction of rest is usually greater 
than that of actual slipping (§ 160). This has been proved 
experimentally in England. The retarding effect of axle and 
rolling friction has been neglected in the above theory. 

Example 1. — A twenty-ton car with an initial velocity of 80 
feet per second (nearly a mile a minute) is to be stopped on a 
level within 1000 feet; required the necessary friction on each 
of the eight wheels. 

Supposing the wheels not to skid, the friction will occur 



FRICTION. 191 

between the brakes and wheels, and is overcome through the 
(relative) distance 1000 feet. Eq. (XYL), § 142, gives (foot- 
lb.-second system) 

0-8FX 1000 = - i ^j( 80 ) 8 > 

from which F ( = friction at circumference of each wheel) 
= 496 lbs. 

Example 2. — Suppose skidding to be impending in the fore- 
going, and the coefficient of friction of rest (i.e., impending 
slipping) between rail and wheel to be f = 0.20. In what 
distance will the car be stopped ? Ans. 496 ft 

Example 3. — Suppose the car in Example 1 to be on an up. 
grade of 60 feet to the mile. (In applying eq. (XYI.) here, 
the weight 20 tons will enter as a resistance.) Ans. 439 lbs. 

Example 4. — In Example 3, consider all four resistances, 
viz., gravity, rolling friction, and brake and axle frictions, the 
distance being 1000 ft., and F the unknown quantity. 

(Take the wheel dimensions of p. 189.) Ans. 414 lbs. 

174. Estimation of Engine and Machinery Friction. — Accord- 
ing to Professor Cotterill, a convenient way of estimating the 
work lost in friction in a steam-engine and machinery driven 
bj it is the following : 

Let jp m = mean effective steam-pressure per unit of area of 
piston, and conceive this composed of three por- 
tions, viz., 
p = the necessary pressure to drive the engine alone un- 
loaded, at the proper speed ; 
p' m — pressure necessary to overcome the resistance caused 

by the useful work of the machines ; 
ejp' m = pressure necessary to overcome the friction of the 
machinery, and that of the engine over and above* 
its friction when unloaded. This is about 15$ of 
p' m (i.e., e = 0.15), except in large engines, and 
then rather less. 
That is, by formula, F being the piston-area and I the length 
of stroke, the work per stroke is thus distributed : 

* Recent experiments (1888) by Prof. Thurston show that this surplus 
engine-friction is practically zero. 



192 MECHANICS OF ENGINEERING. 

p Q is " from 1 to 1-J, or in marine engines 2 lbs. or more per 
square inch." 

175. Anomalies in Friction. — Experiment has shown consid- 
erable deviation under certain circumstances from the laws of 
friction, as stated in § 157 for sliding friction. At pressures 
below f lb. per sq. inch the coefficient f increases when the 
pressure decreases, while above 500 lbs. (Pennie, with iron and 
steel) it increases with the pressure. With high velocities, how- 
ever, above 10 ft. per second, f is much smaller as the velocity 
increases (Bochet, 1858). 

As for axle-friction, experiments instituted bj the Society of 
Mechanical Engineers in England (see the London Engineer 
for March 7 and 21, 1884) gave values for f less than y^- 
when a " bath" of the lubricant was employed. These values 
diminished with increase of pressure, and increased with the 
velocity (see below, Hirn's statement). 

Professor Cotterill says, " It cannot be doubted that for 
values of pv (see § 166) > 5000 the coefficient of friction of 
well-lubricated bearings of good construction diminishes with 
the pressure, and may be much less than the value at low speeds 
as determined by Morin" (p. 259 of his Applied Mechanics). 

Professor Thurston's experiments confirmed those of Hirn as 
to the following relation : " The friction of lubricated surfaces 
is nearly proportional to the square root of the area and pres- 
sure." Hirn also maintained that, "in ordinary machinery, 
friction varies as the square root of the velocity." 

176. Rigidity of Ropes. — If a rope or wire cable passes over 
a pulley or sheave, a force P is required on one side greater 
than the resistance Q on the other for uniform motion, aside 
from axle-friction. Since in a given time both P and Q 
describe the same space s, if P is > Q, then Ps is > Qs, i.e., 
the work done by P is > than that expended upon Q. This 
is because some of the work Ps has been expended in bending 
the stiff rope or cable, and in overcoming friction between the 
strands, both where the rope passes upon and where it leaves 



FRICTION. 



193 



the pulley. With hemp ropes, Fig. 191, the material being 
nearly inelastic, the energy spent in bending it on at D is 
nearly all lost, and energy must also be spent in straightening 




Fig. 191. 

it at E\ but with a wire rope or cable some of this energy is 
restored by the elasticity of the material. The energy spent 
in friction or rubbing of strands, however, is lost in both cases. 
The figure shows geometrically why P must be > Q for a 
uniform motion, for the lever-arm, #, of P is evidently < b 
that of Q. If axle-friction is also considered, we must have 

Pa=Qb+f(P+Q)r, 

r being the radius of the journal. 

Experiments with cordage have been made by Prony, Cou- 
lomb, Eytelwein, and Weisbach, with considerable variation in 
the results and formulae proposed. (See Coxe's translation of 
vol. i., Weisbach's Mechanics.) 

With pulleys of large diameter the effect of rigidity is very 
slight. For instance, Weisbach gives an example of a pulley 
five feet in diameter, with which, Q being = 1200 lbs., P 
= 1219. A wire rope f in. in diameter was used. Of this 
difference, 19 lbs., only 5 lbs. was due to rigidity, the remainder, 
14 lbs., being caused by axle-friction. When a hemp-rope 1.6 
inches in diameter was substituted for the wire one, P— Q=27 
lbs., of which 12 lbs. was due to the rigidity. Hence in one 
case the loss of work was less than -| of Vf , in the other about 
1%, caused by the rigidity. For very small sheaves and thick 
ropes the loss is probably much greater. 
13 



194 MECHANICS OF ENGINEERING. 

177, Miscellaneous Examples. — Example 1. The end of a 
shaft 12 inches in diameter and making 50 revolutions per min- 
ute exerts against its bearing an axial pressure of 10 tons and 
a lateral pressure of 40 tons. With f =f = 0.05, required 
the H. P. lost in friction. Ans. 22.2 H. P. 

Example 2. — A leather belt passes over a vertical pulley, 
covering half its circumference. One end is held by a spring 
balance, which reads 10 lbs. while the other end sustains a 
weight of 20 lbs., the pulley making 100 revolutions per min- 
ute. Required the coefficient of friction, and the H. P. spent 
in overcoming the friction. Also suppose the pulley turned 
in the other direction, the weight remaining the same. The 
diameter of the pulley is 18 inches. A if = 0.22 ; 

US ' (0.142and.2S4H.P. 

Example 3. — A grindstone with a radius of gyration = 12 
inches has been revolving at 120 revolutions per minute, and 
at a given instant is left to the influence of gravity and axle 
friction. The axles are 1-J inches in diameter, and the wheel 
makes 160 revolutions in coming to rest. Required the coeffi- 
cient of axle-friction. Ans. /= 0.389. 

Example 4. — A board A, weight 2 lbs., rests horizontally on 
another B ; coefficient of friction of rest between them being 
f = 0.30. B is now moved horizontally with a uniformly 
accelerated motion, the acceleration being = 15 feet per " square 
second ;" will A keep company with it, or not % Ans. " No." 



PART III. 



STRENGTH OF MATERIALS. 

[Or Mechanics of Materials]. 



CHAPTEB I. 

ELEMENTARY STRESSES AND STRAINS. 

178. Deformation of Solid Bodies. — In the preceding por- 
tions of this work, what was called technically a " rigid 
body," was supposed incapable of changing its form, i.e., 
the positions of its particles relatively to each other, nnder 
the action of any forces to be brought upon it. This sup- 
position was made because the change of form which must 
actually occur does not appreciably alter the distances, 
angles, etc., measured in any one body, among most of 
the pieces of a properly designed structure or machine. 
To show how the individual pieces of such constructions 
should be designed to avoid undesirable deformation or 
injury is the object of this division of Mechanics of En- 
gineering, viz., the Strength of Materials. 




■gr-S 



D 
Fig. 192. § 178. 




As perhaps the simplest instance of the deformation or 
distortion of a solid, let us consider the case of a prismatic 
rod in a state of tension, Fig. 192 (link of a surveyor's 



196 MECHANICS OF ENGINEERING. 

chain, e.g.). The pull at each end is P, and the body is 
said to be under a tension of P (lbs., tons, or other unit), 
not 2P. Let ABGD be the end view of an elementary 
parallelopiped, originally of square section and with faces 
at 45° with the axis of the prism. It is now deformed, the 
four faces perpendicular to the paper being longer* than 
before, while the angles BAD and BGD, originally right 
angles, are now smaller by a certain amount d, ABG and 
ADG larger by an equal amount d. The element is said 
to be in a state of strain, viz.: the elongation of its edges 
(parallel to paper) is called a tensile strain, while the alter- 
ation in the angles between its faces is called a shearing 
strain, or angular distortion (sometimes also called a slid- 
ing, or tangential, strain, since BG has been made to slide, 
relatively to AD, and thereby caused the change of angle). 
[This use of the word strain, to signify change of form and 
not the force producing it, is of recent adoption among 
many, though not all, technical writers.] 

179. Strains. Two Kinds Only. — Just as a curved line may 
be considered to be made up of small straight-line ele- 
ments, so the substance of any solid body may be consid- 
ered to be made up of small contiguous parallelopipeds, 
whose angles are each 90° before the body is subjected to 
the action of forces, but which are not necessarily cubes. 
A line of such elements forming an elementary prism is 
sometimes called & fibre, but this does not necessarily imply 
a fibrous nature in the material in question. The system 
of imaginary cutting surfaces by which the body is thus 
subdivided need not consist entirely of planes ; in the sub- 
ject of Torsion, for instance, the parallelopipedical ele- 
ments considered lie in concentric cylindrical shells, cut 
both by transverse and radial planes. 

Since these elements are taken so small that the only 
possible changes of form in any one of them, as induced 
by a system of external forces acting on the body, are* 

* When a is nearly 0° (or 90°) BG and AD (or AB and DC) are shorter 
than before, on account of lateral contraction; see § 193. 



ELEMENTARY STRESSES, ETC. 197 

elongations or contractions of its edges, and alteration of 
its angles, there are but two kinds of strain, elongation 
(contraction, if negative) and shearing, 

180. Distributed Forces or Stresses. — In the matter preced- 
ing this chapter it has sufficed for practical purposes to 
consider a force as applied at a point of a body, but in 
reality it must be distributed over a definite area ; for 
otherwise the material would be subjected to an infinite 
force per unit of area. (Forces like gravity, magnetic at- 
traction, etc., we have already treated as distributed over 
the mass of a body, but reference is now had particularly 
to the pressure of one body against another, or the action 
of one portion of the body on the remainder.) For in- 
stance, sufficient surface must be provided between the 
end of a loaded beam and the pier on which it rests to 
avoid injury to either. Again, too small a wire must not 
be used to sustain a given load, or the tension per unit 
of area of its cross section becomes sufficient to rupture 
it. 

Stress is distributed force, and its intensity at any point 
of the area is 

>-s ■ • • « 

where dF is a small area containing the point and dP the 
force coming upon that area. If equal dP's (all parallel) 
act on equal dF'soi a plane surface, the stress is said to 
be of uniform intensity, which is then 

*=} • • • • (2) 

where P= total force and i^the total area over which it 
acts. The steam pressure on a piston is an example of 
stress of uniform intensity. 



198 MECHANICS OF ENGINEERING. 

For example, if a force P= 28800 lbs, is uniformly dis- 
tributed over a plane area of F= 72 sq. inches, or ^ of a 
sq. foot, the intensity of the stress is 

28800 Aari „ . ... 

J>= =400 lbs. per sq. men, 

(A 

(or p = 288004-^=57600 lbs. per sq. foot, or p=UA00* 
^=28.8 tons per sq. ft., etc.). 

181. Stresses on an Element; of Two Kinds Only. — When a 
solid body of any material is in equilibrium under a sys- 
tem of forces which do not rupture it, not only is its shape 
altered (i.e. its elements are strained), and stresses pro- 
duced on those planes on which the forces act, but other 
stresses also are induced on some or all internal surfaces 
which separate element from element, (over and above the 
forces with which the elements may have acted on each 
other before the application of the external stresses or 
" applied forces "). So long as the whole solid is the "free 
body " under consideration, these internal stresses, being 
the forces with which the portion on one side of an imag- 
inary cutting plane acts on the portion on the other side, 
do not appear in any equation of equilibrium (for if intro- 
duced they would cancel out); but if we consider free a 
portion only, some or all of whose bounding surfaces are 
cutting planes of the original body, the stresses existing 
on these planes are brought into the equations of equilib- 
rium. 

Similarly, if a single element of the body is treated by 
itself, the stresses on all six of its faces, together with its 
weight, form a balanced system of forces, the body being 
supposed at rest. 




s| 

Fig. 193. 




ELEMENTARY STRESSES, ETC. 199 

As an example of internal stress, consider again the case 
of a rod in tension ; Fig. 193 shows the whole rod (or eye- 
bar) free, the forces P being the pressures of the pins in 
the eyes, and causing external stress (compression here) 
on the surfaces of contact. Conceive a right section made 
through PS, far enough from the eye, G, that we may con- 
sider the internal stress to be uniform in this section, and 
consider the portion PSG as a free body, in Fig. 194. The 
stresses on US, now one of the bounding surfaces of the 
free body, must be parallel to P, i.e., normal to PS; 
(otherwise they would have components perpendicular to 
P, which is precluded by the necessity of 17 being = 0, 
and the supposition of uniformity.) Let F = the sec- 



FlG. 194. 




Fig. 195. 



tional area PS, and p = the stress per unit of area ; then 

IX= gives P= Fp, i,e., p= ? . . (2) 

F 

The state of internal stress, then, is such that on planes 
perpendicular to the axis of the bar the stress is tensile and 
normal (to those planes). Since if a section were made 
oblique to the axis of the bar, the stress would still be 
parallel to the axis for reasons as above, it is evident that 
on an oblique section, the stress has components both nor- 
mal and tangential to the section, the normal component 
being a tension. 



200 



MECHANICS OF ENGINEERING. 



The presence of the tangential or shearing stress in ob- 
lique sections is rendered evident by considering that if an 
oblique dove-tail joint were cut in the rod, Fig. 195, the 
shearing stress on its surfaces may be sufficient to over- 
come friction and cause sliding along the oblique plane. 

If a short prismatic block is under the compressive ac- 
tion of two forces, each = P and applied centrally in one 
base, we may show that the state of internal stress is the 
same as that of the rod under tension, except that the nor- 
mal stresses are of contrary sign, i.e., compressive instead 
of tensile, and that the shearing stresses (or tendency to 
slide) on oblique planes are opposite in direction to those 
in the rod. 

Since the resultant stress on a given internal plane of a 
body is fully represented by its normal and tangential 
components, we are therefore justified in considering but 
two kinds of internal stress, normal or direct, and tangen- 
tial or shearing. 



182. Stress on Oblique Section of Rod in Tension. — Consider 
free a small cubic element whose 
edge =a in length; it has two 
faces parallel to the paper, being 
taken near the middle of the rod 
in Fig. 192. Let the angle which 
the face AB, Fig. 196, makes with 
the axis of the rod be = a. This 
angle, for our present purpose, is 
considered to remain the same 
while the two forces P are acting, 
as before their action. The re- 
sultant stress on the face AB hav- 
ing an intensity p=P-h-F, (see eq. 
2) per unit of transverse section 
of rod, is = p (a sin a) a. Hence 
FlG - m - its component normal to AB is 

pa 2 sin 2 a ; and the tangential or shearing component along 




ELEMENTARY STRESSES, ETC. 201 

AB =pa 2 sin a cos a. Dividing by the area, a 2 , we have 
the following : 

For a rod in simple tension we have, on a plane making 
an angle, a, with the axis : 

a Normal Stress =p sin 2 a per unit of area . . (1) 
and a Shearing Stress =p sin a cos a per unit of area . (2) 

" Unit of area " here refers to the oblique plane in ques- 
tion, while p denotes the normal stress per unit of area of 
a transverse section, i.e., when a=90°, Fig. 194. 

The stresses on CD are the same in value as on AB, 
while for BG and AB we substitute 90° — a for a. Fig. 
197 shows these normal and shearing stresses, and also, 
much exaggerated, the strains or change of form of the 
element (see Fig. 192). 

182a. Relation between Stress and Strain. — Experiment 
shows that so long as the stresses are of such moderate 
value that the piece recovers its original form completely 
when the external forces which induce the stresses are re- 
moved, the following is true and is known as Hooke's Law 
(stress proportional to strain). As the forces P in Fig. 
193 (rod in tension) are gradually increased, the elonga- 
tion, or additional length, of RK increases in the same 
ratio as the normal stress, p } on the sections PS and KN> 
per unit of area [§ 191]. 

As for the distorting effect of shearing stresses, consider 
in Fig. 197 that since 

p sin a cos a = p cos (90° — a) sin (90°— a) 

the shearing stress per unit of area is of equal value C% aU 
four of the faces (perpendicular to paper) in the elementary 
block, and is evidently accountable for the shearing strain, 
i.e., for the angular distortion, or difference, d, between 
90° and the present value of each of the four angles. Ac- 
cording to Hooke's Law then, as P increases within tka 
limit mentioned above, d varies proportionally to 

p sin a cos a, i.e. to the stress. 



202 MECHANICS OF ENGINEERING. 

182b. Example. — Supposing the rod in question were of 
a kind of wood in which a shearing stress of 200 lbs. per 
sq. inch along the grain, or a normal stress of 400 lbs. per 
sq. inch, perpendicular to a fibre-plane will produce rup- 
ture, required the value of a the angle which the grain 
must make with the axis that, as P increases, the danger 
of rupture from each source may be the same. This re- 
quires that 200:400::^ sin a cos a:p sin 2 «, i.e. tan. a must 
= 2.000.*.«=63j^°. If the cross section of the rod is 2 sq. 
inches, the force P at each end necessary to produce rup- 
ture of either kind, when «=63^°, is found by putting 
p sin a cos #= 200. \p =500.0 lbs. per sq. inch. Whence, since 
p=P+F, P=1000 lbs. (Units, inch and pound.) 

183. Elasticity is the name given to the property which 
most materials have, to a certain extent, of regaining their 
original form when the external forces are removed. If 
the state of stress exceeds a certain stage, called the Elastic 
Limit, the recovery of original form on the part of the ele- 
ments is only partial, the permanent deformation being 
called the Set. 

Although theoretically the elastic limit is a perfectly defi- 
nite stage of stress, experimentally it is somewhat indefi- 
nite, and is generally considered to be reached when the 
permanent set becomes well marked as the stresses are in- 
creased and the test piece is given ample time for recovery 
in the intervals of rest. 

The Safe Limit of stress, taken well within the elastic 
limit, determines the working strength or safe load of the 
piece under consideration. E.g., the tables of safe loads 
of the rolled wrought iron beams, for floors, of the New 
Jersey Steel and Iron Co., at Trenton, are computed on 
the theory that the greatest normal stress (tension or com- 
pression) occurring on any internal plane shall not exceed 
12,000 lbs. per sq. inch ; nor the greatest shearing stress 
4,000 lbs. per sq. inch. 



ELEMENTARY STRESSES, ETC. 203 

The tJltimate Limit is reached when rupture occurs. 

184. The Modulus of Elasticity (sometimes called co-efficient 
of elasticity) is the number obtained by dividing the stress 
per unit of area by the corresponding relative strain. 

Thus, a rod of wrought iron y 2 sq. inch sectional area 
being subjected to a tension of 2^ tons =5,000 lbs., it is 
found that a length which was six feet before tension is 
= 6.002 ft. during tension. The relative longitudinal strain 
or elongation is then= (0.002) -r- 6= 1 : 3,000 and the corres- 
ponding stress (being the normal stress on a transverse 
plane) has an intensity of 

p t =P+F= 5,000-^- J^ =10,000 lbs., per sq. inch. 

Hence by definition the modulus of elasticity is (for ten- 
sion) 

# t =_p t -r-e=10,000-T- 3^=30,000,000 lbs. per sq. inch, (the 

sub-script " t " refers to tension). 

It will be noticed that since e is an abstract number, E t 
is of the same quality as p t , i.e., lbs. per sq. inch, or one di- 
mension of force divided by two dimensions of length. 
(In the subject of strength of materials the inch is the 
most convenient English linear unit, when the pound is 
the unit of force ; sometimes the foot and ton are used to- 
gether.) 

The foregoing would be called the modulus of elasticity 
of wrought iron in tension in the direction of the fibre, as 
given by the experiment quoted. But by Hooke's Law p 
and e vary together, for a given direction in a given ma- 
terial, hence within the elastic limit E is constant for a given 
direction in a given material. Experiment confirms this 
approximately. 

Similarly, the modulus of elasticity for compression E 6 



204 MECHANICS OF ENGINEERING. 

in a given direction in a given material may be determined 
by experiments on short blocks, or on rods confined lat- 
erally to prevent flexure. 

As to the modulus of elasticity for shearing, HJ S , we 
divide the shearing stress per unit of area in the given 
direction by d (in n measure) the corresponding angular 
strain or distortion ; e.g., for an angular distortion of 1° or 
<5=.0174, and a shearing stress of 1,566 lbs. per sq. inch, 
we have ^=3^=9,000,000 lbs. per sq. inch. 

Unless otherwise specified, by modulus of elasticity will 
be meant a value derived from experiments conducted 
within the elastic limit, and this, whether for normal stress 
or for shearing, is approximately constant for a given di- 
rection in a given substance.* 

185. Isotropes. — This name is given to materials which 
are homogenous as regards their elastic properties. In 
such a material the moduli of elasticity are individually 
the same for all directions. E.g., a rod of rubber cut out 
of a large mass will exhibit the same elastic behavior when 
subjected to tension, whatever its original position in the 
mass. Fibrous materials like wood and wrought iron are 
not isotropic ; the direction of grain in the former must 
always be considered. The " piling " and welding of nu- 
merous small pieces of iron prevent the resultant forging 
from being isotropic. 

186. Resilience refers to the potential energy stored in a 
body held under external forces in a state of stress which 
does not pass the elastic limit. On its release from con- 
straint, by virtue of its elasticity it can perform a certain 
amount of work called the resilience, depending in amount 
upon the circumstances of each case and the nature of the 
material. See § 148. 

187. General Properties of Materials. — In view of some defi- 
nitions already made we may say that a material is ductile 

* The moduli, or "co-efficients," of elasticity as used by physicists are well explained 
in Stewart and Gee's Practical Physics, Vol. I., pp. 164, etc. Their "co-efficient of 

rigidity" is our E 



ELEMENTARY STRESSES, ETC.~ 205 

when the ultimate limit is far removed from the elastic 
limit ; that it is brittle like glass and cast iron, when those 
limits are near together. A small modulus of elasticity 
means that a comparatively small force is necessary to 
produce a given change of form, and vice versa, but implies 
little or nothing concerning the stress or strain at the 
elastic limit ; thus Weisbach gives E c , lbs. per sq. inch for 
wrought iron = 28,000,000= double the E c for cast iron 
while the compressive stresses at the elastic limit are the 
same for both materials (nearly). 

1887 General Problem of Internal Stress. — This, as treated 
in the mathematical Theory of Elasticity, developed by 
Lame, Clapeyron and Poisson, may be stated as follows : 

Given the original form of a body when free from stress, 
and certain co-efficients depending on its elastic proper- 
ties ; required the new position, the altered shape, and the in- 
tensity of the stress on each of the six faces, of every element 
of the body, when a given balanced system of forces is applied 
to the body. 

Solutions, by this theory, of certain problems of the na- 
ture just given involve elaborate, intricate, and bulky 
analysis ; but for practical purposes Navier's theories 
(1838) and others of more recent date, are sufficiently exact, 
when their moduli are properly determined by experiments 
covering a wide range of cases and materials. These will 
be given in the present work, and are comparatively sim- 
ple. In some cases graphic will be preferred to analytic 
methods as more simple and direct, and indeed for some 
problems they are the only methods yet discovered for ob- 
taining solutions. Again, experiment is relied on almost 
exclusively in dealing with bodies of certain forms under 
peculiar systems of forces, empirical formulae being based 
on the experiments made ; e.g., the collapsing of boilei 
flues, and in some degree the flexure of long columns. 



206 MECHANICS OF ENGINEERING. 

189. Classification of Cases. — Although in almost any case 
whatever of the deformation of a solid body by a balanced 
system of forces acting on it, normal and shearing stresses 
are both developed in every element which is affected at 
all (according to the plane section considered,) still, cases 
where the body is prismatic, and the external system con- 
sists of two equal and opposite forces, one at each end of 
the piece and directed away from each other, are commonly 
called cases of Tension; (Eig. 192); if the piece is a short 
prism with the same two terminal forces directed toward 
each other, the case is said to be one of Compression ; a case 
similar to the last, but where the prism is quite long 
(" long column "), is a case of Flexure or bending, as are also 
most cases where the " applied forces " (i.e., the external 
forces), are not directed along the axis of the piece. Rivet- 
ed joints and " pin-connections " present cases of Shearing; 
a twisted shaft one of Torsion. When the gravity forces 
due to the weights of the elements are also considered, a 
combination of two or more of the foregoing general cases 
may occur. 

In each case, as treated, the principal objects aimed at 
are, so to design the piece or its loading that the greatest 
stress, in whatever element it may occur, shall not exceed 
a safe value ; and sometimes, furthermore, to prevent too 
great deformation on the part of the piece. The first ob- 
ject is to provide sufficient strength; the second sufficient 
stiffness. 

190. Temperature Stresses. — If a piece is under such con- 
straint that it is not free to change its form with changes 
of temperature, external forces are induced, the stresses 
produced by which are called temperature stresses. 



TENSION. 



207 



TENSION. 

191. Hooke's Law by Experiment. — As a typical experiment 
in trie tensiqn of a long rod of ductile metal such as 
wrought iron and the mild steels, the following table is quot- 
ed from Prof. Cotterill's " Applied Mechanics." The experi- 
ment is old, made by Hodgkinson for an English Railway 
Commission, but well adapted to the purpose. From the 
great length of the rod, which was of wrought iron and 
0.517 in. in diameter, the portion whose elongation was 
observed being 49 ft. 2 in. long, the small increase in length 
below the elastic limit was readily measured. The succes- 
sive loads were of such a value that the tensile stress 
p=P+-F, or normal stress per sq. in. in the transverse 
section, was made to increase by equal increments of 2657.5 
lbs. per sq. in., its initial value. After each application of 
load the elongation was measured, and after the removal 
of the load, the permanent set, if any. 



TaUle of elongations of a wrought iron rod, of a lenstb— 49 It. 2 in. 


jp 


X 


jx 


e=X+l 


X 


Load, (lbs. per 
square inch.) 


Elongation, 
(inches.) 


Increment 

of 
Elongation. 


e, the relative 
elongation, (ab- 
stract number.) 


Permanent 

Set, 

(inches.) 


1X2667.5 


.0485 


.0485 


0.000082 




2X " 


.1095 


.061 


.000186 




3X " 


.1675 


.058 


.000283 


0.0015 


4X m 


.224 


.0565 


.000379 


.002 


5X " 


.2805 


.0565 


.000475 


•002T 


6X ** 


.337 


.0565 


.000570 


.003 


7X " 


.393 


.056 




.004 


8X M 


.452 


.059 


.000766 


.0073 


9X M 


.5155 


.0635 




.0195 


10X ■ 


.598 


.0825 




.049 


11 X " 


.760 


.162 




.1545 


12X M 


1.310 


.550 




.667 


etc. 











208 



MECHANICS OF ENGINEERING. 



Referring now to Fig. 198, the notation is evident. P 
is the total load in any experiment, F the cross section of 
the rod ; hence the normal stress on the transverse section 
is p=P- : rF. When the loads are increased by equal in- 
crements, the corresponding increments of the elongation 
X should also be equal if Hooke's law is true. It will be 
noticed in the table that this is very nearly true up to the 
8th loading, i.e., that AX, the difference between two con- 
secutive values of X, is nearly constant. In other words the 
proposition holds good : 



if P and P x are any two loads below the 8th, and X and l x 
the corresponding elongations. 

The permanent set is just perceptible at the 3d load, and 
increases rapidly after the 8th, as also the increment of 
elongation. Hence at the 8th load, which produces a ten- 
sile stress on the cross section of ^9=8x2667.5=21340.0 
lbs. per sq. inch, the elastic limit is reached. 

As to the state of stress of the individual elements, if 

we conceive such sub -division 
of the rod that four edges of 
each element are parallel to the 
axis of the rod, we find that it 
is in equilibrium between two 
normal stresses on its end faces 
w\ (Fi gf 199 ) of a va i ue ^pdF~ 

(P^-F)dF where dF is the hor- 
izontal section of the element. 
If dx was the original length, 
and dX the elongation produced by pdF, we shall have, 
since all the dx'a of the length are equally elongated at the 
same time, 

^ = ; 

dx I 



dz 



Sa 



i 

Fig. 19& 




TENSION. 209 

where h= total (original) length. But dl-^dx is the rela- 
tive elongation e, and by definition (§ 184) the Modulus of 
Elasticity for Tension, E li =p-r-e 

••^=-$7 or *'i3 • • • • (1) 

dx 

Eq. (1) enables us to solve problems involving the elonga- 
tion of a prism under tension, so long as the elastic limit 
is not surpassed. 

The values of E t computed from experiments like those 
just cited should be the same for any load under the elas- 
tic limit, if Hooke's law were accurately obeyed, but in 
reality they differ somewhat, especially if the material 
lacks homogeneity. In the present instance (see Table) 
we have from the 

2d Exper. E=p+e= 28,680,000 lbs. per sq. in. 
5th " E t = " =28,009,000 
8th " E t = " =27,848,000 

If similar computations were made beyond the elastic 
limit, i.e., beyond the 8th Exper., the result would be much 
smaller, showing the material to be yielding much more 
readily. 

192. Strain Diagrams. — If we plot the stresses per sq. inch 
( p) as ordinates of a curve, and the corresponding relative 
elongations (e) as abscissas, we obtain a useful graphic re- 
presentation of the results of experiment. 

Thus, the table of experiments just cited being utilized 
in this way, we obtain on paper a series of points through 
which a smooth curve may be drawn, viz. : OBC Fig. 200, 
for wrought iron. Any convenient scales may be used for 
p and £ ; and experiments having been made on other 
metals in tension and the results plotted to the same scales 



210 



MECHANICS OF ENGINEERING. 



as before for p and e, we have the means of comparing their 
tensile properties. Fig. 200 shows two other curves, rep- 
resenting (roughly) the average behavior of steel and cast 
iron. At the respective elastic limits B, B , and B", it will 
be noticed that the curve for wrought iron makes a sudden 
turn from the vertical, while those of the others curve away 
more gradually ; that the curve for steel lies nearer the 
vertical axis than the others, which indicates a higher 
value for E t ; and that the ordinates BA', B'A', and B'A * 
(respectively 21,000, 9,000, and 30,000 lbs. per sq. inch) in- 




Fig, 200. 

dicate the tensile stress at the elastic limit. These latter 
quantities will be called the moduli of tenacity at elastic 
limit for the respective materials. [On a true scale the 
point G would be much further to the right than here 
shown. Only one half of the curve for steel is given, for 
want of space.] 

Within the elastic limit the curves are nearly straight 
(proving Hooke's law) and if «, a, and a!' are the angles 
made by these straight portions with the axis of X (i.e., 
of e), we shall have 



(E t for w. iron) : (E t c. iron) : (E t steel) : : tan a : tan a' : tan a' 



TENSION. 211 

as a graphic relation between their moduli of elasticity 
(since E=^). 

Beyond the elastic limit the wrought iron rod shows large 
increments of elongation for small increments of stress, 
i.e., the curve becomes nearly parallel to the horizontal 
axis, until rupture occurs at a stress of 53,000 lbs. per sq. 
inch of original sectional area (at rupture this area is some- 
what reduced, especially in the immediate neighborhood 
of the section of rupture ; see next article) and after a rel- 
ative elongation e = about 0.30, or 30%. (The preceding 
table shows only a portion of the results.) The curve 
for steel shows a much higher breaking stress (100,000 
lbs. per sq. in.) than the wrought iron, but the total 
elongation is smaller, e= about 10%. This is an average 
curve ; tool steels give an elongation at rupture of about 
4 to 5%, while soft steels resemble wrought iron in their 
ductility, giving an extreme elongation of from 10 to 20%. 
Their breaking stresses range from 70,000 to 150,000 lbs. 
or more per sq. inch. Cast iron, being comparatively brit- 
tle, reaches at rupture an elongation of only 3 or 4 tenths 
of one per cent., the rupturing stress being about 18,000 
lbs. per sq. inch. The elastic limit is rather ill defined in 
the case of this metal ; and the proportion of carbon and 
the mode of manufacture have much influence on its be- 
havior under test. 

193. Lateral Contraction. — In the stretching of prisms of 
nearly all kinds of material, accompanying the elongation 
of length is found also a diminution of width whose rela- 
tive amount in the case of the three metals just treated is 
about ^ or i^ of the relative elongation (within elastic 
limit). Thus, in the third experiment in the table of § 191, 
this relative lateral contraction or decrease of diameter 
= ^to^of e, i.e., about 0.00008. In the case of cast 
iron and hard steels contraction is not noticeable ex- 



212 



MECHANICS OF ENGINEERING 



cept by very delicate measurements, both within and with* 
out the elastic limit ; but the more ductile metals, as 
wrought iron and the soft steels, when stretched beyond 
the elastic limit show this feature of their deformation 
in a very marked degree. Fig. 201 shows by dotted lines 
the original contour of a wrought iron rod, while the con- 
tinuous lines indicate that at rupture. At the cross section 
of rupture, whose position is determined by some 
local weakness, the drawing out is peculiarly 
pronounced. 

The contraction of area thus produced is some- 
times as great as 50 or 60% at the fracture. 

194. "Flow of Solids." — When the change in re- 
lative position of the elements of a solid is ex- 
treme, as occurs in the making of lead pipe, 
drawing of wire, the stretching of a rod of duc- 
tile metal as in the preceding article, we have 
Fi&. 201. instances of what is called the Floiv of Solids, in- 
teresting experiments on which have been made by 
Tresca. 




195. Moduli of Tenacity. — The tensile stress per square 
inch (of original sectional area) required to rupture a 
prism of a given material will be denoted by T and called 
the modulus of ultimate tenacity ; similarly, the modulus oj 
safe tenacity, or greatest safe tensile stress on an element, 
by T' ; while the tensile stress at elastic limit may be 
called T". The ratio of T' to T" is not fixed in practice 
but depends upon circumstances (from ^ to ^). 

Hence, if a prism of any material sustains a total pull 
or load P, and has a sectional area=i^, we have 



P= FT for the ultimate or breaking load. 

p=FT' " " safe load. 

p"= FT" " " load at elastic limit. 



(2) 



Of course T' should always be less than T". 



TENSION. 



213 



196. Resilience of a Stretched Prism. — Fig. 202. In the 
gradual stretching of a prism, fixed at one extremity, the 
value of the tensile force P at the other necessarily de- 
pends on the elongation X at each stage of the lengthening, 
according to the relation [eq. (1) of § 191.] 



PI 



FE t 



(3) 



within the elastic limit. (If we place a weight G on the 
flanges of the unstretched prism and then leave 
it to the action of gravity and the elastic action 
of the prism, the weight begins to sink, meeting 
an increasing pressure P, proportional to X, from 
the flanges). Suppose the stretching to continue 
until P reaches some value P" (at elastic limit 
Ja say), and X a value X". Then the work done so 
Y far is 

fig. 802. ^ = mean force x space = y 2 P" X" . . (4) 
But from (2) P=FT , and (see §§ 184 and 191) 




A"=e"Z 
■\ (4) becomes U=}4 T e". Fl=% T" e" V 



(5) 



where Vis the volume of the prism. The quantity }4T"e" t 
or work done in stretching to the elastic limit a cubic 
inch (or other unit of volume) of the given material, Weis- 
bach calls the Modulus of Resilience for tension. From (5) 
it appears that the amounts of work done in stretching to 
the elastic limit prisms of the same material but of differ- 
ent dimensions are proportional to their volumes simply. 
The quantity }4T"e" is graphically represented by the 
area of one of the triangles such as OA'B, 0A"B" in Fig. 
200 ; for (in the curve for wrought iron for instance) the 
modulus of tenacity at elastic limit is represented by A'B, 
and e" (i.e., e for elastic limit) by OA'. The remainder of 



214 MECHANICS OF ENGINEERING. 

the area OBG included between the curve and the hori- 
zontal axis, i.e., from B to G, represents the work done in 
stretching a cubic unit from the elastic limit to the point 
of rupture, for each vertical strip having an altitude =p 
and a width = de, has an area =pde, i.e., the work done by 
the stress p on one face of a cubic unit through the dis- 
tance de, or increment of elongation. 

If a weight or load = G be " suddenly "applied to stretch 
the prism, i.e., placed on the flanges, barely touching 
them, and then allowed to fall, when it comes to rest again 
it has fallen through a height X ly and experiences at this 
instant some pressure P l from the flanges ; P x =? The 
work Gh has been entirely expended in stretching the 
prism, none in changing the kinetic energy of G, which 
=0 at both beginning and end of the distance l l7 

.'. GK-V2PA ..P^ZG. 

Since P 1 =2G f i.e., is > G, the weight does not remain in 
this position but is pulled upward by the elasticity of the 
prism. In fact, the motion is harmonic (see §§59 and 
138). Theoretically, the elastic limit not being passed, the 
oscillations should continue indefinitely. 

Hence a load G " suddenly applied " occasions double the 
tension it would if compelled to sink gradually by a sup- 
port underneath, which is not removed until the tension is 
just = G, oscillation being thus prevented. 

If the weight G sinks through a height =h before strik- 
ing the flanges, Fig. 202, we shall have similarly, within 
elastic limit, if ^= greatest elongation, (the mass of rod 
being small compared with that of G). 

Oih+Jft-.tfPA .... (6) 

If the elastic limit is to be just reached we have from eqs. 
(5) and (6), neglecting ^ compared with h, 

Gh=y 2 T"e"V ... (7) 



TENSION. 



215 



an equation of condition that the prism shall not be in- 
jured. 

Example. — If a steel prism have a sectional area of y^ 
eq. inch and a length Z=10 ft. =120 inches, what is the 
greatest allowable height of fall of a weight of 200 lbs., 
that the final tensile stress induced may not exceed T"= 
30,000 lbs. per sq. inch, if e" =.002 ? From (7), using the 
inch and pound, we have 

^ ! ^__ 30,000x.002x^xl20 =45 inches> 
2G ~ 2x200 

197. Stretching of a Prism by Its Own Weight. — In the case 
of a very long prism such as a mining- 
pump rod, its weight must be taken into 
account as well as that of the terminal 
load P„ see Fig. 203. At (a.) the prism 
is shown in its unstrained condition ; at 
(b) strained by the load P x and its own 
weight. Let the cross section be = F, the 
heaviness of the prism =y. Then the rela- 
tive extension of any element at a distance 




<$ from o is 



_dl {P.+rFx) 



dx 



FE t 



(1) 



(See eq. (1) § 191) ; since P L +Fyx is the load hanging upon 
the cross section at that locality. Equal cfcc's, therefore, 
are unequally elongated, x varying from to I. The total 
elongation is 



a= rut= _L r 

Jo FU t Jo 



1 [Pjx+rFxdx] = H+H? 



Le., X= the amount due to P 1} plus an extension which 
half the weight of the prism would produce, hung at the 
lower extremity. 



216 MECHANICS OE ENGINEERING. 

The foregoing relates to the deformation of the piece, 
and is therefore a problem of stiffness. As to the strength 
of the prism, the relative elongation e=dX- : rdx [see eq. (1)], 
which is variable, must nowhere exceed a safe value e'= 
T'+Bt (from eq. (1) § 191, putting P=FT\ and X=X). 
Now the greatest value of the ratio dX : dx, by inspecting 
eq. (1), is seen to be at the upper end where x =1. The 
proper cross section F, for a given load P u is thus found. 

Putting £d&?} = 2L we have F =^— 1 . (2) 

8 FF t E,' T'—yl v ' 

198. Solid of Uniform Strength in Tension, or hanging body 
of minimum material supporting its own 
weight and a terminal load P x . Let it be a 
solid of revolution. If every cross-section 
F at a distance =x from the lower extrem- 
ity, bears its safe load FT', every element 
of the body is doing full duty, and its form 
is the most economical of material. 

The lowest section must have an area 
Fi«. 204. Fo==Pi + T ^ gince Pi is its safe loa(L rig# 

204. Consider any horizontal lamina ; its weight is yFdx, 
(j= heaviness of the material, supposed homogenous), and 
its lower base F must have P x -\- G for its safe load, i.e. 

G+P X =FT' . . . (1) 

in which G denotes the weight of. the portion of the solid 
below F. Similarly for the upper base F-\-dF 9 we have 

G+P^yFdx^F+dF^' . . (2) 

By subtraction we obtain 

rFdx=T'dF; i.e. ?dx= *£ 
T' F 




TENSION. 217 

in which the two variables x and F are separated. By in- 
tegration we now have 

J F 

\x P, yx 

Le., F=F er =±± er (4) 

from which i^may be computed for any value of x. 

The weight of the portion below any F is found from (1) 
and (4) ; i.e. 

G=P 1 {^—l}» (5) 

while the total extension X will be 



*='"~l (6) 

the relative elongation dX-r-dx being the same for every dx 
and bearing the same ratio to e" (at elastic limit), as T' 
does to T". 

199. Tensile Stresses Induced by Temperature. — If the two 
ends of a prism are immovably fixed, when under no strain 
and at a temperature t, and the temperature is then low- 
ered to a value t\ the body suffers a tension proportional 
to the fall in temperature (within elastic limit). If for a 
rise or fall of 1° Fahr. (or Cent.) a unit of length of the 
material would change in length by an amount rj (called 
the co-efficient of expansion) a length =1 would be con- 
tracted an amount X=-rjl(t-t') during the given fall of tem- 
perature if one end were free. Hence, if this contraction 
is prevented by fixing both ends, the rod must be under a 
tension P, equal in value to the force which would be 



218 MECHANICS OF ENGINEERING. 

necessary to produce the elongation X, just stated, under 
ordinary circumstances at the lower temperature. 

From eq. (1) §191, therefore, we have for this tension 
due to fall of temperature 



P= ^l(t—t')=E t F(t--t')7 j 



For 1° Cent, we may write 

For Cast iron y = .0000111 ; 
" Wrought iron = .0000120 ; 
" Steel = .0000108 to .0000114 ; 

" Copper y = .0000172 ; 

" Zinc r) = .0000300. 



COMPRESSION OF SHORT BLOCKS. 

200. Short and Long Columns. — In a prism in tension, its 
own weight being neglected, all the elements between thi 
localities of application of the pair of external forces pro- 
ducing the stretching are in the same state of stress, if the 
external forces act axially (excepting the few elements in the 
immediate neighborhood of the forces ; these suffering 
local stresses dependent on the manner of application of 
the external forces), and the prism may be of any length 
without vitiating this statement. But if the two external 
forces are directed toward each other the intervening ele- 
ments will not all be in the same state of compressive 
stress unless the prism is comparatively short (or unless 
numerous points of lateral support are provided). A long 
prism will buckle out sideways, thus even inducing tensile 
stress, in some cases, in the elements on the convex side. 

Hence the distinction between short Hocks and long 
columns. Under compression the former yield by crush- 
ing or splitting, while the latter give way by flexure (i.e. 
bending). Long columns, then will be treated separately 



COMPRESSION OF SHORT BLOCKS. 219 

in a subsequent chapter. In the present section the blocks 
treated being about three or four times as long as wide, 
all the elements will be considered as being under equal 
compressive stresses at the same time. 

201. Notation for Compression. — By using a subscript c, 
we may write 

E Q = Modulus of Elasticity;* i.e. the quotient of the 
compressive stress per unit of area divided by the relative 
shortening ; also 

G— Modulus of crushing ; i.e. the force per unit of sec- 
tional area necessary to rupture the block by crushing ; 

(7'= Modulus of safe compression, a safe compressive 
stress per unit of area ; and 

G"= Modulus of compression at elastic limit. 

For the absolute and relative shortening in length we 
may still use i ande, respectively, and within the elastic 
limit may write equations similar to those for tension, F 
being the sectional area of the block and P one of the ter- 
minal forces, while p = compressive stress per unit of area 
of F, viz.: 

w _p _ P+F ^ P+F _Pl m 

c 7 di^tic ~mj fx • : • ' K ' 

within the elastic limit. 
Also for a short block 

Crushing force =FG \ 

Compressive force at elastic limit —FG" > . (2) 
Safe compressive force =FG f ) 

202. Remarks on Crushing. — As in § 182 for a tensile 
stress, so for a compressive stress we may prove that a 

* [Note. — It must be remembered that the modulus of elasticity, 
whether for normal or shearing stresses, is a number indicative of stiff- 
ness, not of strength, and has no relation to the elastic limit (except 
that experiments to determine it must not pass that limit).] 



220 MECHANICS OF ENGINEERING. 

shearing stress —p sin a cos a is produced on planes at an 
angle a with the axis of the short block, p being the com- 
pression per unit of area of transverse section. Accord- 
ingly it is found that short blocks of many comparatively 
brittle materials yield by shearing on planes making an 
angle of about 45° with the axis, the expression p sin a 
cos a reaching a maximum, for «=45° ; that is, wedge- 
shaped pieces are forced out from the sides. Hence the 
necessity of making the block three or four times as long 
as wide, since otherwise the friction on the ends would 
cause the piece to show a greater resistance by hindering 
this lateral motion. Crushing by splitting into pieces 
parallel to the axis sometimes occurs. 

Blocks of ductile material, however, yield by swelling 
out, or bulging, laterally, resembling plastic bodies some- 
what in this respect. 

The elastic limit is more difficult to locate than in ten- 
sion, but seems to have a position corresponding to that 
in tension, in the case of wrought iron and steel. With 
cast iron, however, the relative compression at elastic 
limit is about double the relative extension (at elastic 
limit in tension), but the force producing it is also double. 
For all three metals it is found that U c =B t quite nearly, 
so that the single symbol i7may be used for both. 



EXAMPLES IN TENSION AND COMPRESSION. 

203. Tables for Tension and Compression. — The round num- 
bers of the following tables are to be taken as rude averages 
only, for use in the numerical examples following. (The 
scope and design of the present work admit of nothing 
more. For abundant detail of the results of the more im- 
portant experiments of late years, the student is referred 
to the recent works of Profs. Thurston, Burr, Lanza, and 
Wood). Another column might have been added giving 
the Modulus of Eesilience in each case, viz.: y 2 e"T" 



(which also =__ ) ; see § 196. e is an abstract 



num- 



EXAMPLES IN TENSION AND COMPRESSION. 221 

ber, and =X-±-l, while E t , T", and T are given in pounds 
per square inch: 



TABLE OF 


THE MODULI, ETC 


., OF MATERIALS IN TENSION, 






i" 


£ 


% 


rp II 


T 


Material. 


(Elastic limit. ) 


At Rupture. 


Mod. of Elast. 


Elastic limit. 


Rupture. 




abst. number. 


abst. number. 


lbs. per sq. in. 


lbs. per sq. in. 


lbs. per sq. in. 


Soft Steel, 




.00200 


.2500 


26,000,000 


50,000 


80,000 


Hard Steel, 




.00200 


.0500 


40,000,000 


90,000 


130,000 


Cast Iron, 




.00066 


.0020 


14,000,000 


9,000 


18,000 
I 45,000 
< to 
( 60,000 

16,000 
to 

50,000 


Wro't Iron, 
Brass, 




.00080 
.00100 


.2500 


28,000,000 
10,000,000 


22,000 

( 7,000 
\ to 
( 19,000 


Glass, 








9,000,000 




3,500 


Wood, with 
the fibres, 


1 


.00200 

to 
.01100 


.0070 

to 

.0150 


200,000 
to 
2,000,000 


3,000 

to 
19,000 


6,000 

to 
28,000 


Hemp rope, 












7,000 



[N.B. — Expressed in kilograms per square centim., E^, T and T" would be nu 
merically about 1 / li as large as above, while e and e" would be unchanged.] 



TABLE OF MODULI, ETC.; COMPRESSION OF SHORT BLOCKS. 





e" 


£ 


E e 


G" 


G 


Material. 


Elastic limit. 


At lupture. 


Mod. of Elast. 


Elastic limit. 


Rupture. 




abst. number 


abst. number. 


lbs. per sq. in. 


lbs. per sq. in. 


lbs. per sq. in. 


Soft Steel, 


0.00100 




30,000,000 


30,000 




Hard Steel. 


0.00120 


0.3000 


40,000,000 


50,000 


200,000 


Cast Iron, 


0.00150 




14,000,000 


20,000 


90,000 


Wro't Iron, 


0.00080 


0.3000 


28,000,000 


24,000 


40,000 


Glass, 










20,000 


Granite, 

Sandstone, 

Brick, 






1 See 
J §213a 




10,000 
5,000 
3,000 


Wood, with 
the fibres, 




( 0.0100 
< to 
.( 0.0400 


350,000 

to 

2,000,000 




2,000 

to 
10,000 


Portland 1 
Cement, f 






(§ 213a) 




4,000 



222 MECHANICS OF ENGINEERING. 

204. Examples. No. 1. — A bar of tool steel, of sectional 
area =0.097 sq. inches, is ruptured by a tensile force of 
14,000 lbs. A portion of its length, originally y 2 a foot, 
is now found to have a length of 0.532 ft. Required T, 
and s at rupture. Using the inch and pound as units (as 
in the foregoing tables) we have T= iig?= 144326 lbs. per 
sq. in.; (eq. (2) § 195) ; while 

e=(0.532—0.5) x 12-^(0.50 X 12)=0.064. 

Example 2. — Tensile test of a bar of " Hay Steel " for 
the Glasgow Bridge, Missouri. The portion measured was 
originally 3.21 ft. long and 2.09 in. X 1.10 in. in section. 
At the elastic limit P was 124,200 lbs., and the elongation 
was 0.064 ins. Required E ti T', and e" (for elastic limit). 

e''=-i= ^° 6 t =-00165 at elastic limit. 
I 3.21x12 

2*"= 124,200--(2.09x 1.10)= 54,000 lbs. per sq. in. 

^=f=£=2#xlS65= 32 ' 570 ' 000 lb8 - *» "* in - 

Nearly the same result for E t would probably have been 
obtained for values of p and e below the elastic limit. 

The Modulus of Besilience of the above steel (see § 196) 
would be y 2 e" T"= 44.82 inch -pounds of work per cubic 
inch of metal, so that the whole work expended in stretch- 
ing to the elastic limit the portion above cited is 

£7= y 2 e " T" T=3968. inch-lbs. 

An equal amount of work will be done by the rod in re- 
covering its original length. 

Example 3. — A hard steel rod of *4 sq. in. section and 
20 ft. long is under no stress at a temperature of 130 c 



EXAMPLES IN TENSION AND COMPRESSION. 223 

Cent., and is provided with flanges so that the slightest 
contraction of length will tend to bring two walls nearer 
together. If the resistance to this motion is 10 tons how 
low must the temperature fall to cause any motion ? y be- 
ing = .0000120 (Cent, scale). From § 199 we have, ex- 
pressing P in lbs. and F in sq. inches, since E t = 40,000,000 
lbs. per sq. inch, 

10x2,000=40,000,000 X }4 X (130-*') X 0.000012; whence 
£'=46.6° Centigrade. 

Example 4. — If the ends of an iron beam bearing 5 tons 
at its middle rest upon stone piers, required the necessary 
bearing surface at each pier, putting C for stone = 200 
lbs. per sq. inch. 25 sq. in., Ans. 

Example 5. — How long must a wrought iron rod be, 
supported vertically at its upper end, to break with its 
own weight ? 216,000 inches, Ans. 

Example 6. — One voussoir (or block) of an arch -ring 
presses its neighbor with a force of 50 tons, the joint hav- 
ing a surface of 5 sq. feet ; required the compression per 
sq. inch. 138.8 lbs. per sq. in., Ans. 

205. Factor of Safety. — -When, as in the case of stone, the 
value of the stress at the elastic limit is of very uncertain 
determination by experiment, it is customary to refer the 
value of the safe stress to that of the ultimate by making 
it the w'th portion of the latter, n is called a factor of 
safety, and should be taken large enough to make the safe 
stress come within the elastic limit. For stone, n should 
not be less than 10, i,e. C'—G~?-n\ (see Ex. 6, just given), 



206. Practical Notes. — It was discovered independently by 
Commander Beardslee and Prof. Thurston, in 1873, that 
if wrought iron rods were strained considerably beyond 
the elastic limit and allowed to remain free from stress 



224 MECHANICS OF ENGEtfEEBLCTGk 

for at least one day thereafter, a second test would show 
higher limits both elastic and ultimate. 

When articles of cast iron are imbedded in oxide of iron 
and subjected to a red heat for some days, the metal loses 
most of its carbon, and is thus nearly converted into 
wrought iron, lacking, however, the property of welding., 
geing malleable, it is called malleable cast iron. 

Chrome steel (iron and chromium) and tungsten steel pos- 
sess peculiar hardness, fitting them for cutting tools, rock 
drills, picks, etc. 

B y fatigue of metals we understand the fact, recently dis- 
covered by Wohler in experiments made for the Prussian 
Grovernment, that rupture may be produced by causing the 
stress on the elements to vary repeatedly between two- 
limiting values, the highest of which may be considerably 
below T (or (7), the number of repetitions necessary to- 
produce rupture being dependent both on the range of 
variation and the higher value. 

For example, in the case of Phoenix iron in tension,, 
rupture was produced by causing the stress to vary from 
) to 52,800 lbs. per sq. inch, 800 times ; also, from to 
14,000 lbs. per sq. inch 240,853 times ; while 4,000,000 va- 
riations between 26,400 and 48,400 per sq. inch did not 
cause rupture. Many other experiments were made and 
the following conclusions drawn (among others): 

Unlimited repetitions of variations of stress (lbs. per 
#q. in.) between the limits given below will not injure the 
metal (Prof. Burr's Materials of Engineering), 

w , . . f From 17,600 Comp. to 17,600 Tension. 

Wrought rron. { .„ * Q ^ ^ „ 

( From 30,800 Comp. to 30,800 Tension 
Axle Cast Steel.-] " to 52,800 

( " 38500 Tens, to 88,000 
(See p. 232 for an addendum to this paragraph.) 



SHEAKING. 

SHEARING. 



225 



207. Rivets. — The angular distortion called shearing 
strain in the elements of a body, is specially to be provided 
for in the case of rivets joining two or more plates. This 
distortion is shown, in Figs. 205 and 206, in the elements 
near ';he plane of contact of the plates, much exaggerated. 




-1 



V\\ 



- ? 2P 

3 — 



\J7 



M. 



Fig. 205. 



Fig, 206. 



In Fig. 205 (a lap-joint) the rivet is said to be in single 
shear ; in Fig. 206 in double shear. If P is just great 
enough to shear off the rivet, the modulus of ultimate shear- 
ing, which may be called S, (being the shearing force per 
unit of section when rupture occurs) is 



8- 



F iTtd 2 



(1) 



in which F= the cross section of the rivet, its diameter 
being =d. For safety a value S'= J^ to yi of S should 
be taken for metal, in order to be within the elastic limit. 

As the width of the plate is diminished by the rivet 
hole the remaining sectional area of the plate should be 
ample to sustain the tension P, or 2P, (according to the 
plate considered, see Fig. 206), P being the safe shearing 
force for the rivet. Also the thickness t of the plate 
should be such that the side of the hole shall be secure 
against crushing ; P must not be > C'td, Fig. 205. 

Again, the distance a, Fig. 205, should be such as to 
prevent the tearing or shearing out of the part of the 
plate between the rivet and edge of the plate. 



226 



MECHANICS OF ENGINEERING. 



For economy of material the seam or joint should be 
no more liable to rupture by one than by another, of the 




Fig. 207. 

four modes just mentioned. The relations which must 
then subsist will be illustrated in the case of the " butt- 
joint " with two cover -plates, Fig. 207. Let the dimen- 
sions be denoted as in the figure and the total tensile force 
on the joint be = Q. Each rivet (see also Fig. 206) is ex- 
posed in each of two of its sections to a shear of i 2 Q, 
hence for safety against shearing of rivets we put 



kQ-%^s 



(1) 



Along one row of rivets in the main plate the sectional 
area for resisting tension is reduced to (b — 3d)t l} hence for 
safety against rupture of that plate by the tension Q, we 
put 



Q^b—Zd^T' 



(2) 



Equations (1) and (2) suffice to determine d for the rivets 
and t Y for the main plates, Q and b being given; but the 
values thus obtained should also be examined with refer- 
ence to the compression in the side of the rivet hole, i.e. 5 
y 6 Q must not be > G%d. [The distance a, Fig. 205, to the 
edge of the plate is recommended by different authorities 
to be from d to 3d] 

Similarly, for the cover -plate we must have 



and } 2 Q not >C'td, 



y 2 Qor(b—U)tT' 
< 



. (3) 



SHEARING. 227 

If the rivets do not fit their holes closely, a large margin 
should be allowed in practice. Again, in boiler work, the 
pitch, or distance between centers of two consecutive rivets 
may need to be smaller, to make the joint steam-tight, than 
would be required for strength alone, 

208. Shearing Distortion. — The change of form in an ele- 
ment due to shearing is an angular deformation and will 
be measured in tt -measure. This angular change or dif- 
ference between the value of the corner angle during strain 
and }4x, its value before strain, will be called d, and is 
proportional (within elastic limit) to the shearing stress 
per unit of area, J9 S , existing on all the four faces whose 
angles with each other have been changed. 

Fig. 208. (See § 181). By § 184 the Modulus of Shearing 
Elasticity is the quotient obtained by dividing p s by d ; i.e. 
(elastic limit not passed), 

^ s =^ . . . . (1) 
or inversely, S=p s ~E s (1)' 

The value of E s for different substances is most easily 
determined by experiments on torsion 
in which shearing is the most promi- 
nent stress. (This prominence depends 
on the position of the bounding planes 
of the element considered ; e.g., in Fig. 
208, if another element were considered 
-dx—\ within the one there shown and with 

fig. 208. its planes at 45° with those of the first, 

we should find tension alone on one pair of opposite faces, 
compression alone on the other pair.) It will be noticed 
that shearing stress cannot be present on two opposite 
faces only, but exists also on another pair of faces (those 
perpendicular to the stress on the first), forming a couple 
of equal and opposite moment to the first, this being 
necessary for the equilibrium of the element, even when 




228 



MECHANICS OF ENGINEERING. 



tensile or compressive stresses are also present on the 
faces considered. 

209. Shearing Stress is Always of the Same Intensity on the 
Four Faces of an Element. — (By intensity is meant per unit 
of area ; and the four faces referred to are those perpen- 
dicular to the paper in Fig. 208, the shearing stress being 
parallel to the paper.) 

Let dx and dz be the width and height of the element 
in Fig. 208, while dy is its thickness perpendicular to the 
paper. Let the intensity of the shear on the right hand 
face be =(/ R , that on the top face =p s . Then for the ele- 
ment a» a free body, taking moments about the axis per- 
pendicular to paper, we have 

q a dz dy X dx — p s dx dy X dz =0 .*. q 8 =p 8 
(dx and dz being the respective lever arms of the forces 
q s dz dy and p s dx dy.) 

Even if there were also tensions (or compressions) on 
one or both pairs of faces their moments about would 
balance (or fail to do so by a differential of a higher order) 
independently of the shears, and the above result would 
still hold. 



210. Table of Moduli for Shearing. 





d" 


E t 


S" 


8 


Material. 


i.e. 5 at elastic 
limit. 


Mod. of Elasticity 
for Shearing. 


(Elastic limit.) 


(Rupture.) 




arc in 7r-measure. 


lbs. per sq. in. 


lbs. per sq. in 


lbs. per sq. in. 


Soft Steel, 




9,000,000 




70,000 


Hard Steel, 


0.0032 


14,000,000 


45,000 


90,000 


Cast Iron, 


0.0021 


7,000,000 


15,000 


30,000 


Wrought Iron, 


0.0022 


9,000,000 


20,000 


50,000 


Brass, 




5,000,000 






Glass, 










Wood, across I 
fibre, 1 








1,500 

to 

8,000 


Wood, along ( 








500 

to 

1,200 


fibre, 1 









SHEARING. 



229 



As in the tables for tension and compression, the above 
values are averages. The true values may differ from 
these as much as 30 per cent, in particular cases, accord' 
ing to the quality of the specimen. 

211. Punching rivet holes in plates of metal requires the 
overcoming of the shearing resistance along the convex 
surface of the cylinder punched out. Hence if d = diam- 
eter of hole, and t = the thickness of the plate, the neces- 
sary force for the punching, the surface sheared being 
F= tnd, is 



P=St7:d 



(2)' 



Another example of shearing action is the " stripping " 
of the threads of a screw, when the nut is forced off lon- 
gitudinally without turning, and resembles punching in 
its nature. 

212. E and E s ; Theoretical Relation. — In case a rod is in 
tension within the elastic limit, the relative (linear) lateral 
contraction (let this =ra) is so connected with E t and E s 
that if two of the three are known the third can be de- 
duced theoretically. This relation is proved as follows, 
by Prof. Burr. Taking an elemental cube with four of its 
faces at 45° with the axis of the piece, Fig. 209, the axial 
half-diagonal AD becomes of a length AD'=AD-\-e.AD 
under stress, while the transverse half diagonal contracts 
to a length B'D'=AD—m.AD, The angular distortion d 





Fig. 209. § 212. 



Fig. 210. 



230 MECHANICS OF ENGINEERING. 

is supposed very small compared with 90° and is due to 
the shear p s per unit of area on the face BG (or BA). 
From the figure we have 

, tAKo d, B'D' 1— m -. 

tan(45°— _) = __=___ =l_ m — s, approx. 

[But, Fig. 210, tan(45° — x)=l — 2a? nearly, where a; is a 
small angle, for, taking CA= unity= A E, tan AD=AF= 
AE—EF. Now approximately EF=^G.^/2siudEG= 
BD^/2=x^/2 .: AF= l—2x nearly.] Hence 

1— d=l— m— e; or d=m-{-e . . (2) 

Eq. (2) holds good whatever the stresses producing the 
deformation, but in the present case of a rod in tension, 
if it is an isotrope, and if p = tension per unit of area on 
its transverse section, (see § 182, putting «=45°), we have 
E t =p-±-e and i7 s =( p s on BG)-i-o= yip-^o. Putting also 
(ra :e)=r, whence m=re, eq. (2) may finally be written 

m Hr+1) k ;U ' E °=mh) ■ ■ (3) 

Prof. Bauschinger, experimenting with cast iron rods, 
found that in tension the ratio m : £ was = S, as an average, 
which in eq. (3) gives 

^=-^^=1^ nearly. . . . (4) 

His experiments on the torsion of cast iron rods gave 
E s = 6,000,000 to 7,000,000 lbs. per sq. inch. By (4), then, 
E t should be 15,000,000 to 17,500,000 which is approxi- 
mately true (§ 203). 

Corresponding results may be obtained for short blocks 
in compression, the lateral change being a dilatation in- 
stead of a contraction. 




SHEARING. 231 

£13. Examples in Shearing. — Example 1. — Bequired the 
proper length, a, Fig. 211, to 
guard against the shearing off, 
along the grain, of the portion 
ab, of a wooden tie-rod, the force 
P being = 2 tons, and the width 
of the tie = 4 inches. Using a 
value of S' = 100 lbs. per sq. in., 
we put haS'*= 4,000 cos 45° ; i.e. 
Rb. 211. a= (4,000 x 0.707) -K4x 100)= 7.07 

inches. 

Example 2. — A 7/% in. rivet of wrought iron, in single 
shear (see Eig. 205) has an ultimate shearing strength 
P=FS=}{7:d 2 S=y7r{^fx50 ) 000=30,050lbs. For safety, 
putting S'= 8,000 instead of #,P'= 4,800 lbs. is its safe 
shearing strength in single shear. 

The wrought iron plate, to be secure against the side- 
crushing in the hole, should have a thickness t, computed 
thus : 

P'=tdC'\ or 4,800=^x12,000 .-. £=0.46 in. 

If the plate were only 0.23 in. thick the safe value of P 
would be only y 2 of 4,800. 

Example 3. — Conversely, given a lap-joint, Fig. 205, in 
which the plates are y in. thick and the tensile force on 
the joint = 600 lbs. per linear inch of seam, how closely 
must y^ inch rivets be spaced in one row, putting #'=8,000 
and O =12,000 lbs. per sq. in.? Let the distance between 
centres of rivets be =x (in inches), then the force upon 
each rivet =600#, while its section i<=0.44sq. in. Having 
regard to the shearing strength of the rivet we put 600x= 
0.44x8,000 and obtain x— 5. 8 6 in.; but considering that the 
safe crushing resistance of the hole is = 14^. 12,000= 
2,250 lbs., 600^=2,250 gives ^=3.75 inches, which is the 
pitch to be adopted. What is the tensile strength of the 
reduced sectional area of the plate, with this pitch ? 



232 MECHANICS OF ENGINEERING. 

Example 4. — Double butt-joint ; (see Fig. 207) ; % inch 
plate; ^ in. rivets; 2"= C'=12,000 ; #'=8,333; width of 
plates =14 inches. Will one row of rivets be sufficient at 
each side of joint, if §=30,000 lbs.? The number of rivets 
= ? Here each rivet is in double shear and has therefore 
a double strength as regards shear. In double shear the 
safe strength of each rivet =2FS'= 7,333 lbs. Now 30,000-- 
7,333=4.0 (say). With the four rivets in one row the re- 
duced sectional area of the main plate is =[14 — 4x ^] X 3 /s 
=4.12 sq. in., whose safe tensile strength is =jF 7 T=4.12x 
12,000=49,440 lbs.; which is > 30,000 lbs. .-. main plate is 
safe in this respect. But as to side-crushing in holes 
in main plate we find that G%d (i.e. 12,000 X 3 / 8 X % =3,375 
lbs.) is <%Q i.e.<7,500 lbs., the actual force on side of 
hole. Hence four rivets in one row are too few unless 
thickness of main plate be doubled. Will eight in one 
row be safe ? 

213a. (Addendum to § 206.) Elasticity of Stone and Cements. 
— Experiments by Gen. Gillmore with the large Watertown 
testing-machine in 1883 resulted as follows (see p. 221 for 
notation) : 

With cubes of Haverstraw Freestone (a homogeneous brown- 
stone) from 1 in. to 12 in. on the edge, E was found to be 
from 900,000 to 1,000,000 lbs. per sq. in. approximately ; and 
C about 4,000 or 5,000 lbs. per sq. in. Cubes of the same 
range of sizes of Dyckerman's Portland cement gave E c from 
1,350,000 to 1,630,000, and C from 4,000 to 7,000, lbs. per sq. 
in. Cubes of concrete of the above sizes, made with the 
Newark Cc.'s Posendale cement, gave E c about 538,000, while 
cubes of cement-mortar, and some of concrete, both made with 
National Portland cement, showed E c from 800,000 to 2,000,- 
000 lbs. per sq. in. 

The compressibility of brick piers 12 in. square in section 
and 16 in. high was also tested. They were made of common 
North Piver brick with mortar joints f in. thick, and showed 
a value for E c of about 300,000 or 400,000, while at elastic 
limit C" was on the average 1,000, lbs. per sq. in. 



TOBSION. 



233 



CHAPTEE II. 

TORSION. 

214. Angle of Torsion and of Helix. When a cylindrical 
beam or shaft is subjected to a twisting or torsional action, 
i. e. when it is the means of holding in equilibrium two 
couples in parallel planes and of equal and opposite mo- 
ments, the longitudinal axis of symmetry remains straight 

and the elements along it exper- 
ience no stress (whence it may be 
Icalled the "line of no twist"), 
while the lines originally parallel to 
it assume the form of helices, each 
element of which is distorted in its angles (originally 
right angles), the amount of distortion being assumed pro- 
portional to the radius of the helix. The directions of the 





Fig. 213. 

faces of any element were originally as follows : two radial, 
two in consecutive transverse sections, and the other two 
tangent to two consecutive circular cylinders whose com- 
mon axis is that of the shaft. E.g. in Fig. 212 we have 
an unstrained shaft, while in Fig. 213 it holds the two 



234 



MECHANICS OF ENGINEERING. 



couples (of equal moment P a — Q b) in equilibrium. These 
couples act in parallel planes perpendicular to the axis of 
the prism and a distance, Z, apart. Assuming that the 
transverse sections remain plane and parallel during tor- 
sion, any surface element, m, which in Fig. 212 was entire- 
ly right-angled, is now distorted. Two of its angles have 
been increased, two diminished, by an amount d f the angle 
between the helix and a line parallel to the axis. Suppos- 
ing m to be the most distant of any element from the axis, 
this distance being e, any other element at a distance z 



from the axis experiences an angular distortion — ■ 



o. 



If now we draw B' parallel to O'A the angle B B' t 
=a, is called the Angle of Torsion, while d may be called the 
helix angle; the former lies in a transverse plane, the latter 
in a plane tangent to the cylinder. Now 

tan d = (linear arc B B')-?-l; but lin. arc B B' = ea ; hence, 
putting d for tan d, (d being small) 



ea 
T 



(1) 



(d and a both in n measure). 

215. Shearing Stress on the Elements. The angular distor- 
tion, or shearing strain, d> of any element (bounded as al- 
ready described) is due to the shearing stresses exerted on 
it by its neighbors on the four faces perpendicular to the 

tangent plane of the cylindri- 
cal shell in which the element 
is situated. Consider these 
neighboring elements of an 
outside element removed, and 
the stresses put in ; the latter 
are accountable for the dis- 
tortion of the element and so 




Fig. 214. 



TOBSIOtf. 235 

hold it in equilibrium. Fig. 214 shows this element 
"free." Within the elastic limit d is known to be propor- 
tional to p s , the shearing stress per unit of area on the 
faces whose relative angular positions have been changed. 
That is, from eq. (1) § 208, o=p s - : rE s ; whence, see (1) of 

§ 214, 

*■— n (2) 

In (2) p s and e both refer to a surface element, e being 
the radius of the cylinder, and p s the greatest intensity of 
shearing stress existing in the shaft. Elements lying nearer 
the axis suffer shearing stresses of less intensity in pro- 
portion to their radial distances, i.e., to their helix-angles. 
That is, the shearing stress on that face of the element 
which forms a part of a transverse section and whose dis- 
tance from the axis is s, is p, =— p si per unit of area, and 

e 

the total shear on the face is pdF, dF being the area of the 
face. 



216. Torsional Strength. — We are now ready to expose the 
full transverse section of a shaft under torsion, to deduce 
formulae of practical utility. Making a right section of 
the shaft of Fig. 213 anywhere between the two couples 
and considering the left hand portion as a free body, the 
forces holding it in equilibrium are the two forces P of 
the left-hand couple and an infinite number of shearing 
forces, each tangent to its circle of radius z, on the cross 
section exposed by the removal of the right-hand portion. 
The cross section is assumed to remain plane during tor- 
sion, and is composed of an infinite number of dF's, each 
being the area of an exposed face of an element ; see Fig. 
215. 



236 



MECHANICS OF ENGINEERING. 




Fig. 215. 



Each elementary shearing force = J. p R dF, and % is its 
lever arm about the axis Oo . For equilibrium, 2 (mom.) 
about the axis Oo must =0 ; i.e. in detail 

— py 2 ar-py 2 a+ f ( S p B dF)z=0 



or, reducing, 



^ 8 fz*dF=Pa; or, Ml=P« 
eJ e 



(3) 



Eq. (3) relates to torsional strength, since it contains p s , the 
greatest shearing stress induced by the torsional couple, 
whose moment Pa is called the Moment of Torsion, the 
stresses in the cross section forming a couple of equal and 
opposite moment. 

7 P is recognized as the Polar Moment of Inertia of the cross 
section, discussed in § 94 ; e is the radial distance of the 
outermost element, and = the radius for a circular shaft- 

217. Torsional Stiffness. — In problems involving the angle 
of torsion, or deformation of the shaft, we need an equa- 
tion connecting Pa and a, which is obtained by substitut- 
ing in eq. (3) the value of p s in eq. (2), whence 



i 



(4) 



torsion. 237 

From this it appears that the angle of torsion, a, is propor- 
tional to the moment of torsion, Pa, within the elastic 
limit ; a must be expressed in it-measure, Trautwine cites 1° 
(i.e. a= 0.0174) as a maximum allowable value for shafts. 

218. Torsional Resilience is the work done in twisting a 
shaft from an unstrained state until the elastic limit is 
reached in the outermost elements. If in Fig. 213 we 
imagine the right-hand extremity to be fixed, while the 
other end is gradually twisted through an angle a Y each 
force P of the couple must be made to increase gradually 
from a zero value up to the value P lf corresponding to a,. 
In this motion each end of the arm a describes a space 
= ]/ 2 aa x , and the mean value of the force = %P\ (compare 
§ 196). Hence the work done in twisting is 

U l =}4PiX}£aa 1 x2=)4P l aa 1 . . (5) 
By the aid of preceding equations, (5) can be written 

rr-^m or - p ' w at—nW m 

If for p s we write S' (Modulus of safe shearing) we have 
for the safe resilience of the shaft 

U'=-^L . . . . (7) 

If the torsional elasticity of an originally unstrained shaft 
is to be the means of arresting the motion of a moving 
mass whose weight is G, (large compared with the parts 
intervening) and velocity =v 9 we write (§ 133) 

9 2 
as the condition that the shaft shall not be injured. 



238 MECHANICS OF ENGINEERING. 

219. Polar Moment of Inertia. — For a shaft of circular 
cross section (see § 94) 7 p =i^^r 4 ; for a hollow cylinder 
i p = j47r(r^~r 2 ^) ; while for a square shaft I p =y^b i , b being 
the side of the square ; for a rectangular cross-section 
sides b and h, I v =lJbJi(l} 2 -\-h 2 ). For a cylinder e=r; if hol- 
low, e=r , the greater radius. For a square, e=J^&^/2. 

220. Non-Circular Shafts. — If the cross-section is not cir- 
cular it becomes warped, in torsion, instead of remaining 
plane. Hence the foregoing theory does not strictly ap- 
ply. The celebrated investigations of St. Yenant, how- 
ever, cover many of these cases. (See § 708 of Thompson 
and Tait's Natural Philosophy ; also, Prof. Burr's Elas- 
ticity and Strength of the Materials of Engineering). His 
results give for a square shaft (instead of the 

Pa of eq. (4) of § 217), 



61 



Pa=0Ml ^1 . . . . (1) 

Do 



and Pa=V 5 6 3 j9 s , instead of eq. (3) of § 216, p s being the 
greatest shearing stress. 

The elements under greatest shearing strain are found 
at the middles of the sides, instead of at the corners, when 
the prism is of square or rectangular cross-section. The 
warping of the cross-section in such a case is easily veri- 
fied by the student by twisting a bar of india-rubber in 
his fingers. 

221. Transmission of Power. — Fig. 216. Suppose the cog- 
wheel B to cause A, on the 
same shaft, to revolve uni- 
formly and overcome a resis- 
tance Q, the pressure of the 
teeth of another cog-wheel, 
i-P B being driven by still another 
fig. 216. wheel. The shaft AB is un- 




TORSION. 239 



der torsion, the moment of torsion being = Pa= Qb. (P x 
and Qi the hearing reactions have no moment about the 
axis of the shaft). If the shaft makes u revolutions per 
Tinit-time, the work transmitted [transmitted ; not expend- 
ed in twisting the shaft whose angle of torsion remains 
constant, corresponding to Pa) per unit-time, i.e. the Power, 
is 

L=P.27ra.u=27ruPa . . . (8) 

To reduce L to Horse Power (§ 132), we divide by N 9 
the number of units of work per unit-time constituting 
one H. P. in the system of units employed, i.e., 

Horse Power =H. P.= 2?z 



For example i\r=33,000 ft, -lbs. per minute, or =396,000 
inch -lbs. per minute ; or = 550 ft.-lbs. per second. Usually 
the rate of rotation of a shaft is given in revolutions per 
minute, 

But eq. (8) happens to contain Pa the moment of torsion 
acting to maintain the constant value of the angle of tor- 
sion, and since for safety (see eq. (3) § 216) Pa=S'I p -r-e, 
with 7 P = Y^tlt 4 ' and e=r for a solid circular shaft, we have 
ior such a shaft 

(Safe),aP,^y ... (9) 

which is the safe H. P., which the given shaft can trans- 
mit at the given speed. B' may be made 7,000 lbs. per sq. 
inch for wrought iron ; 10,000 for steel, and 5,000 for cast- 
iron. If the value of Pa fluctuates periodically, as when 
a shaft is driven by a connecting rod and crank, for (H. P.) 
we put raX(H. P.), m being the ratio of the maximum to 
the mean torsional moment ; m = about V/ 2 under ordi- 
nary circumstances (Cotterill). 



240 



MECHANICS OF ENGINEERING. 



222. Autographic Testing Machine. — The principle of Prof 
Thurston's invention bearing this name is shown in Fig 




Fie. 217. 



217. The test-piece is of a standard shape and size, its 
central cylinder being subjected to torsion. A jaw, carry- 
ing a handle (or gear-wheel turned by a worm) and a drum 
on which paper is wrapped, takes a firm hold of one end 
of the test-piece, whose further end lies in another jaw 
rigidly connected with a heavy pendulum carrying a pen- 
cil free to move axially. By a continuous slow motion of 
the handle the pendulum is gradually deviated more and 
more from the vertical, through the intervention of the 
test-piece, which is thus subjected to an increasing tor- 
sional moment. The axis of the test-piece lies in the axis 
of motion. This motion of the pendulum by means of a 
properly curved guide, WR, causes an axial (i.e., parallel 
to axis of test-piece) motion of the pencil A, as well as an 
angular deviation /3 equal to that of the pendulum, and 
this axial distance CF,=sT, of the pencil from its initial 
position measures the moment of torsion =_Pa=I > c sin /?. 
As the piece twists, the drum and paper move relatively 
to the pencil through an angle sUo equal to the angle 



TOKSIOK 



241 



of torsion a so far attained. The abscissa so and ordinate 
s T of the cnrve thus marked on the paper, measure, 
when the paper is unrolled, the values of a and Pa through 
all the stages of the torsion. Fig. 218 shows typical 




Fig. 218. 

curves thus obtained. Many valuable indications are 
given by these strain diagrams as to homogeneousness of 
composition, ductility, etc., etc. On relaxing the strain 
at any stage within the elastic limit, the pencil retraces 
its path ; but if beyond that limit, a new path is taken 
called an "elasticity-line," in general parallel to the first 
part of the line, and showing the amount of angular re- 
covery, BC, and the permanent angular set, OB. 

223. Examples in Torsion.— The modulus of safe shearing 
strength S', as given in § 221, is .expressed in pounds per 
square inch ; hence these two units should be adopted 
throughout in any numerical examples where one of the 
above values for S' is used. The same statement applies 
to the modulus of shearing elasticity, E s , in the table of 
§ 210. 

Example 1.— Fig. 216. With P = 1 ton, a = 3 ft, I = 
10 ft, and the radius of the cylindrical shaft r=2.5 inches, 
required the max. shearing stress per sq. inch, p s , the 
shaft being of wrought iron. From eq. (3) § 216 

Pa e __ 2,000x36x2.5 . qqnl , . 

~, V^X(2^ 2 »930 lbs. per sq. inch, 

which is a safe value for any ferrous metal. 



242 



MECHANICS OF ENGINEERING. 



Example 2. — What H. P. is the shaft in Ex. 1 transmit- 
ting, if it makes 50 revolutions per minute ? Let u = 
number of revolutions per unit of time, and N ' = the num- 
ber of units of work per unit of time constituting one 
horse-power. Then H. ~P.=Pu27ia~N i which for the foot- 
pound-minute system of units gives 

H. P.=2,000x50x27rx3-r-33,000=57^ H. P. 



Example 3. — What different radius should be given to 
the shaft in Ex. 1, if two radii at its extremities, originally 
parallel, are to make an angle of 2° when the given moment 
of torsion is acting, the strains in the shaft remaining con- 
stant. From eq. (4) § 217, and the table 210, with «= 1 g ^= 
0.035 radians (i.e. x -measure), and 7 p = 1 / 2 7rr 4 , we have 



2,000x36x120 



^0.035x9,000,000 



-=17.45 .-. r=2.04 inches. 



(This would bring about a different p s , but still safe.) The 
foregoing is an example in stiffness. 

Example 4. — A working shaft of steel (solid) is to trans- 
mit 4,000 H. P. and make 60 rev. per minute, the maximum 
twisting moment being 1% times the average; required 
its diameter. d=14.74 inches. Ans. 

Example 5. — In example 1, p,= 2,930 lbs. per square 
inch ; what tensile stress does this imply on a plane at 45° 
with the pair of planes on which p s acts ? Eig. 219 shows 



[ Ps d 1 ** 



p,dz 





Fig. 220. 



TORSION. 243 

a small cube, of edge =dx, (taken from the outer helix of 
Fig. 215,) free and in equilibrium, the plane of the paper 
being tangent to the cylinder ; while 220 shows the portion 
BD G, also free, with the unknown total tensile stresses 2 ^/2 
acting on the newly exposed rectangle of area =dxXdx^/2 } 
p being the unknown stress per unit of area. From sym- 
metry the stress on this diagonal plane has no shearing 
component. Putting ^[components normal to BD]=0, 
we have 

pdx 2 */2=2dx 2 p ii cos4:5 =dx 2 p s ^/2.:p=p s . (1) 

That is, a normal tensile stress exists in the diagonal 
plane BD of the cubical element equal in intensity to the 
shearing stress on one of the faces, i.e., =2,930 lbs. per sq. 
in. in this case. 

Similarly in the plane A G will be found a compressive 
stress of 2,930 lbs. per sq. in. If a plane surface had been 
exposed making any other angle than 45° with the face of 
the cube in Fig. 219, we should have found shearing and 
normal stresses each less than p a per sq. inch. Hence the 
interior dotted cube in 219, if shown " free " is in tension 
in one direction, in compression in the other, and with 
no shear, these normal stresses having equal intensities. 
Since S' is usually less than T' or G', if p s is made = S' 
the tensile and compressive actions are not injurious. It 
follows therefore that when a cylinder is in torsion any 
helix at an angle of 45° with the axis is a line of tensile, 
or of compressive stress, according as it is a right or left 
handed helix, or vice versa. 

Example 6. — A solid and a hollow cylindrical shaft, of 
equal length, contain the same amount of the same kind 
of metal, the solid one fitting the hollow of the other. 

Compare their torsional strengths, used separately. 
The solid shaft has only ¥L the strength of the hollow 
one. Ans. 



244 MECHANICS OF ENGKLtfEERIKGh 



CHAPTER III. 

FLEXURE OF HOMOGENEOUS PRISMS UNDER 
PERPENDICULAR FORCES IN ONE PLANE. 

224. Assumptions of the Common Theory of Flexure. — When 
a prism is bent, under the action of external forces per- 
pendicular to it and in the same plane with each other, it 
may be assumed that the longitudinal fibres are in tension 
on the convex side, in compression on the concave side, 
and that the relative stretching or contraction of the ele- 
ments is proportional to their distances from a plane in- 
termediate between, with the understanding that the flex- 
ure is slight and that the elastic limit is not passed in any 
element. 

This " common theory " is sufficiently exact for ordinary 
engineering purposes if the constants employed are prop- 
erly determined by a wide range of experiments, and in- 
volves certain assumptions of as simple a nature as possi- 
ble, consistently with practical facts. These assumptions 
are as follows, (for prisms, and for solids with variable cross 
sections, when the cross sections are similarly situated as 
regards a central straight axis) and are approximately 
borne out by experiment : 

(1.) The external or " applied " forces are all perpendicu- 
lar to the axis of the piece and lie in one plane, which may 
be called the force-plane ; the force-plane contains the 
axis of the piece and cuts each cross-section symmetri- 
cally ; 

(2.) The cross-sections remain plane surfaces during 
flexure ; 

(3.) There is a surface (or, rather, sheet of elements) 
which is parallel to the axis and perpendicular to the 
force-plane, and along which the elements of the solid ex- 



FLEXUEE. 



245 



perience no tension nor compression in an axial direction, 
this being called the Neutral Surface; 

f4.) The projection of the neutral surface upon the force 
plane (or a || plane) being called the Neutral Line or Elastic 
Curve, the bending or flexure of the piece is so slight that 
an elementary division, ds, of the neutral line may be put 
=dx, its projection on a line parallel to the direction of 
the axis before flexure ; 

(5.) The elements of the body contained between any 
two consecutive cross-sections, whose intersections with 
the neutral surface are the respective Neutral Axes of the 
sections, experience elongations (or contractions, accord- 
ing as they are situated on one side or the other of the 
neutral surface), in an axial direction, whose amounts are 
proportional to their distances from the neutral axis, and 
indicate corresponding tensile or compressive stresses ; 

(6.) E t =E ; 

(7.) The dimensions of the cross-section are small com- 
pared with the length of the piece ; 

(8.) There is no shear perpendicular to the force plane 
on internal surfaces perpendicular to that plane. 

In the locality where any one of the external forces is 
applied, local stresses are of course induced which demand 
separate treatment. These are not considered at present. 

225. Illustration. — Consider the case of flexure shown in 
Fig. 221. The external forces are three (neglecting the 




Fig. 221. 



246 



MECHANICS OF ENGINEERING. 



weight of the beam), viz.: P l9 P 2 , and P 3 . P x and P 3 are 
loads, P 2 the reaction of the support. 

The force plane is vertical. N Y L is the neutral line or 
elastic curve. NA is the neutral axis of the cross-section 
at m ; this cross-section, originally perpendicular to the 
sides of the prism, is during flexure ~| to their tangent 
planes drawn at the intersection lines ; in other words, the 
side view QNB 9 of any cross-section is perpendicular to 
the neutral line. In considering the whole prism free we 
have the system P l9 P 2 , and P 3 in equilibrium, whence 
from IY=0 we have P 2 =P 1 -\rP^ 9 and from ^(mom. about 
0) =0, P 3 ? 3 =P 1 £ 1 . Hence given P l we may determine the 
other two external forces. A reaction such as P 2 is some- 
times called a supporting force. The elements above the 
neutral surface N x 0LS are in tension ; those below in com- 
pression (in an axial direction). 

226. The Elastic Forces. — Conceive the beam in Fig. 221 
separated into two parts by any transverse section such 
as QA 9 and the portion NiON 9 considered as a free body 
in Fig. 222. Of this free body the surface QAB is one of 




eOsTic 



"*■ l/ <r 




fiG. 222. 



FLEXURE. 247 

the bounding surfaces, but was originally an internal sur- 
face of the beam m Fig. 221. Hence in Fig. 222 we must 
put in the stresses acting on all the dF's or elements of area 
of QAB. These stresses represent the actions of the body 
taken away upon the body which is left, and according to 
assumptions (5), (6) and (8) consist of normal stresses (ten- 
sion or compression) proportional per unit of area, to the 
distance, z, of the dF's from the neutral axis, and of shear- 
ing stresses parallel to the force-plane (which in most 
cases will be vertical). 

The intensity of this shearing stress on any dF varies 
with the position of the dF with respect to the neutral 
axis, but the law of its variation will be investigated later 
(§§ 253 and 254). These stresses, called the Elastic Forces 
of the cross-section exposed, and the external forces P x and 
P 2 , form a system in equilibrium. We may therefore ap- 
ply any of the conditions of equilibrium proved in § 38. 

227. The Neutral Axis Contains the Centre of Gravity of the 
Cross-Section. — Fig. 222. Let e.= the distance of the outer- 
most element of the cross-section from the neutral axis, and 
the normal stress per unit of area upon it be =jo, whether 
tension or compression. Then by assumptions (5) and (6), 
§ 224, the intensity of normal stress on any dF is = -1 p 
and the actual 

normal stress on any dF is=~ pdF . (1) 

This equation is true for dF's having negative z's, i.e. 
on the other side of the neutral axis, the negative value 
of the force indicating normal stress of the opposite char- 
acter ; for if the relative elongation (or contraction) of two 
axial fibres is the same for equal s's, one above, the other 
below, the neutral surface, the stresses producing the 
changes in length are also the same, provided F t =F c ; see §§ 
184 and 201. 



248 MECHANICS OF ENGINEERING. 

For this free body in equilibrium put JX=0 (X is a 
horizontal axis). Put the normal stresses equal to their 
X components, the flexure being so slight, and the X com- 
ponent of the shears = for the same reason. This gives 
(see eq. (1) ) 

C± pdF= ; i.e. JL PdFz= ; or, S. Fz=0 (2) 

la which z= distance of the centre of gravity of the cross - 
section from the neutral axis, from which, though un- 
known in position, the z's have been measured (see eq. 
(4) § 23). 

In eq. (2) neither p-^-e nor F can be zero .\ z must = ; 
i.e. the neutral axis contains the centre of gravity. Q. E. D. 
[If the external forces were not all perpendicular to the 
beam this result would not be obtained, necessarily.] 

228. The Shear. — The " total shear," or simply the 
"shear," in the cross-section is the sum of thje vertical 
shearing stresses on the respective dF's. Call this sum 
J 9 and we shall have from the free body in Fig. 222, by 
putting 2Y=Q (Y being vertical) 

F 2 —F l —J=0r.J=F 2 —P 1 . . (3) 

That is, the shear equals the algebraic sum of the ex- 
ternal forces acting on one side (only) of the section con- 
sidered. This result implies nothing concerning its mode 
of distribution over the section. 

229. The Moment— By the "Moment of Flexure" or 
simply the Moment, at any cross- section is meant the sum 
of the moments of the elastic forces of the section, taking 
the neutral axis as an axis of moments. In this summa- 
tion the normal stresses appear alone, the shear taking no 
part, having no lever arm about the neutral axis. Hence, 
Fig. 222, the moment of flexure 



FLEXURE. 249 

This function, CdFz 2 , of the cross-section or plane figure 

is the quantity called Moment of Inertia of a plane figure, 
§ 85. For the free body in Fig. 222, by putting ^(mom.s 
about the neutral axis NA)=0, we have then 



2— — P l x i -\-P 2 x 2 =0 > or in general. S- =31 . (5) 
e e 



in which 31 signifies the sum of moments, about the neutral 
axis of the section, of all the forces acting on the free body 
considered, exclusive of the elastic forces of the exposed 
section itself. 



230. Strength, in Elexure. — Eq. (5) is available for solving 
problems involving the Strength of beams and girders, since 
it contains p, the greatest normal stress per unit of area to 
be found in the section. 

In the cases of the present chapter, where all the exter- 
nal forces are perpendicular to the prism or beam, and 
have therefore no components parallel to the beam, i.e. to 
the axis X, it is evident that the normal stresses in any 
section, as QB Fig. 222, are equivalent to a couple ; for the 
condition -X=0 falls entirely upon them and cannot be 
true unless the resultant of the tensions is equal, parallel, 
and opposite to that of the compressions. These two equal 
and parallel resultants, not being in the same line, form a 
couple (§ 28), which we may call the stress-couple. The 
moment of this couple is the " moment of flexure " p ~ , and 
it is further evident that the remaining forces in Fig. 222, 
viz.: the shear J and the external forces P t and P 2 , are 
equivalent to a couple of equal and opposite moment to 
the one formed by the normal stresses. 



250 



MECHANICS OF ENGINEERING. 



231. Flexural Stiffness.— The neutral line, or elastic curve, 
containing the centres of gravity of all the sections, was 
originally straight ; its radius of curvature at any point, 
as N, Tig. 222, during flexure may be introduced as fol- 
lows. QB and U'V are two consecutive cross-sections, 
originally parallel, but now inclined so that the intersec- 
tion (7, found by prolonging them sufficiently, is the centre 
of curvature of the da (put =dx) which separates them at 
N, and CG= t o= the radius of curvature of the elastic 
curve at N. From the similar triangles TJ'TJG and GNGwe 
have dk :dx::e: p, in which dX is the elongation, IT Z7, of a 
portion, originally =dx, of the outer fibre. But the rela- 
tive elongation s=— of the latter is, by §184, within the 

elastic limit, =M??*JL 
E E 



P 



and eq. (5) becomes 



EI 



=M 



(6) 



From (6) the radius of curvature can be computed. E= 
the value of E t =E c , as ascertained from experiments in 
bending. 

To obtain a differential equation of the elastic curve, (6) 
may be transformed thus, Fig. 223. The curve being very 

flat, consider two consecutive 
da's with equal dx's; they may 
be put = their dx's. Produce 
the first to intersect the dy of the 
second, thus cutting off the d 2 y, 
i.e. the difference between two 



AXIS X 



ax 




ft>%J, 

90 ° <$ % y cons ecutive dy's. Drawing a per- 
pendicular to each da at its left 
extremity, the centre of curva- 
ture C is determined by their in- 
tersection, and thus the radius 
of curvature p. The two shaded 
FlQ - 223 - triangles have their small angles 



FLEXURE. 2M 

equal, and d?y is nearly perpendicular to the prolonged 
ds ; hence, considering them similar, we have 

P :dx::dx:d?y.:~=^ 

and hence from eq. (6) we have 

(approx.) ±EI*jL=M ' ' W 

as a differential equation of the elastic curve. From this 
the equation of the elastic curve may be found, the de- 
flections at different points computed, and an idea thus 
formed of the stiffness. All beams in the present chap- 
ter being prismatic and lumogeneous both E and /are the 
same (i.e. constant) at all points of the elastic curve. In 
using (7) the axis Xmust be taken parallel to the length 
of the beam before flexure, which must be slight ; the 
minus sign in (7) provides for the case when (Py+dx* is es- 
sentially negative. 

232 Resilience of Flexure.— If the external forces are made 
to increase gradually from zero up to certain maximum 
values, some of them may do work, by reason of their 
points of application moving through certain distances 
due to the yielding, or flexure, of the body. If at the be- 
ginning and also at the end of this operation the body is 
at rest, this work has been expended on the elastic resis- 
tance of the body, and an equal amount, called the work 
of resilience (or springing-back), will be restored by the 
elasticity of the body, if released from the external forces, 
provided the elastic limit has not been passed. The energy 
thus temporarily stored is of the potential kind; see §§ 
148, 180, 196 and 218. 

232a. Distinction Between Simple, and Continuous, Beams (or 
"Girders").— The external forces acting on a beam consist 



252 



MECHANICS OF ENGINEERING. 



generally of the loads and the " reactions " of the sup= 
ports. If the beam is horizontal and rests on two supports 
only, the reactions of those supports are easily found by 
elementary statics [§ 36] alone, without calling into ac- 
count the theory of flexure, and the beam is said to be a 
Simple Beam, or girder ; whereas if it is in contact with 
more than two supports, being " continuous," therefore, 
over some of them, it is a Continuous Girder (§ 271). The 
remainder of this chapter will deal only with simple 
beams. 



ELASTIC CURVES, 



233. Case I. Horizontal Prismatic Beam, , Supported at Both 
Ends, With a Central Load, Weight of Beam Neglected. — Fig. 
224. First considering the whole beam free, we find each 



-H7- 

—\ — 



^ 



Fig. 224. § 



^ 



reaction to be =%P. AOB is the neutral line ; required 
the eqiiation of the portion OB referred to as an origin, 
and to the tangent line through as the axis of X To 
do this consider as free the portion mB between any sec- 
tion m on the right of and the near support, in Fig. 
225. The forces holding this free body in equilibrium 



By^Otl 




Fig. 225. 



Fig. 226. 



ELASTIC CURVES. 253 

are the one external force ^P, and the elastic forces act- 
ing on the exposed surface. The latter consist of J", the 
shear, and the tensions and compressions represented in 
the figure by their equivalent " stress-couple." Selecting 
JST, the neutral axis of m, as an axis of moments (that J 
may not appear in the moment equation) and putting 
2 (mom) =0 we have 

2 V 2 ) dx 2 dx 2 2 \ 2 / 



Fig. 226 shows the elastic curve OB in its purely geomet- 
rical aspect, much exaggerated. For axes and origin as in 
figure d 2 y^-dx 2 is positive. 

Eq. (1) gives the second cc-derivative of y equal to a 
function of x. Hence the first ^-derivative of y will be 
equal to the #-anti-derivative of that function, plus a con- 
stant, C. (By anti-derivative is meant the converse of de- 
rivative, sometimes called integral though not in the sense 
of summation). Hence from (1) we have (EI being a con- 
stant factor remaining undisturbed) 



EI 



dy P ( I x 2 * 



a £=T(k*-%)+ G • ' ™ 



(2)' is an equation between two variables dy-^-dx and x, and 
holds good for any point between and B; dys-dx de- 
noting the tang, of «, the slope, or angle between the tan- 
gent line and X At the slope is zero, and x also zero ; 
hence at (2/ becomes 

#1x0=0— 0+C 

which enables us to determine the constant 0, whose value 
must be the same at as for all points of the curve. 
Hence C=Q and (2)' becomes 



254 



MECHANICS OF ENGENDERING. 



FI±= F -(lx-t\ 
dx 2 \ 2 2) 



. (2) 



from which the slope, tan. a, (or simply «, in iz -measure ; 
since the angle is small) may be found at any point. Thus 
at B we have x=yil and dy-t-dx=a lf and 

_1 PI 2 
•'• ai ~16 * EI 

Again, taking the #-anti-derivative of both members of eq. 
(2) we have 



P /te* x? 



^-f(x-?) +(7 ' 



(3)' 



and since at both x and y are zero, G' is zero. Hence 
the equation of the elastic curve OB is 



Ely 



<S-S) 



(31 



To compute the deflection of from the right line join- 
ing A and .Bin Fig. 224, i.e. BK, =d, we put x=%l in (3), y 
being then =d, and obtain 



BK=d= 



48 * EI 



(4) 



Eq. (3) does not admit of negative values for x ; for if 
the free body of Fig. 225 extended to the left of 0, the ex< 
ternal forces acting would be P, downward, at ; and y&P, 
upward, at B, instead of the latter alone ; thus altering 
the form of eq. (1). From symmetry, however, we know 
that the curve AO, Fig. 224, is symmetrical with OB about 
the vertical through 0. 



ELASTIC CURVES. 255 

233a. Load Suddenly Applied. — Eq. (4) gives the deflection 
d corresponding to the force or pressure P applied at the 
middle of the beam, and is seen to be proportional to it. 
If a load G hangs at rest from the middle of the beam, 
P=G; but if the load G, being initially placed at rest 
upon the unbent beam, is suddenly released from the ex- 
ternal constraint necessary to hold it there, it sinks and 
deflects the beam, the pressure P actually felt by the beam 
varying with the deflection as the load sinks. What is 
the ultimate deflection d m ? Let P m = the pressure be- 
tween the load and the beam at the instant of maximum 
deflection. The work so far done in bending the beam 
= j4P m d m . The potential energy given up by the load 
= Gd m , while the initial and final kinetic energies are both 
nothing. 

-% Gd m =y 2m Pd m . . (5) 

That is, P m = 2G. Since at this instant the load is sub- 
jected to an upward force of 2G and to a downward force 
of only G (gravity) it immediately begins an upward mo- 
tion, reaching the point whence the motion began, and 
thus the oscillation continues. We here suppose the elas- 
ticity of the beam unimpaired. This is called the " sud- 
den " application of a load, and produces, as shown above, 
double the pressure on the beam which it does when grad- 
ually applied, and a double deflection. The work done 
by the beam in raising the weight again is called its re- 
silience. 

Similarly, if the weight G is allowed to fall on the mid- 
dle of the beam from a height h, we shall have 

Gx(h+dJ, or approx., Gh=}^P m d m ; 

and hence, since (4) gives d m in terms of P mi 

96 EI ; 1? K ' 



256 



MECHANICS OF ENGINEERING. 




Fig 22? 



This theory supposes the mass of the beam small com- 
pared with the falling weight. 

234. Case II. Horizontal Prismatic Beam, Supported at Both 
Ends, Bearing a Single Eccentric Load. Weight of Beam Neg- 
lected. — Fig. 227. The reactions 
of the points of support, P and 
Pj, are easily found by consider- 
ing the whole beam free, and put- 
ting first Jfmom.J^O, whence P l 
=Pl-t-l lt and then l(m.om..) B =O y 
whence P =P(l l — l)-r-l lm P and 
P 1 will now be treated as known quantities. 

The elastic curves OC and CB, though having a common 
tangent line at C (and hence the same slope « c ), and a com- 
mon ordinate at C, have separate equations and are both 
referred to the same origin and axes, as shown in the 
figure. The slope at 0, «o> an( i that at B,a l9 are unknown 
constants, to be determined in the progress of the work. 

Equation of OC. — Considering as free a portion of the 
beam extending from B to a section made anywhere on 
OC, x and y being the co-ordinates of the neutral axis of 
that section, we conceive the elastic forces put in on the 
exposed surface, as in the preceding problem, and put 
Jfrnom. about neutral axis of the section) =0 which gives 
(remembering that here d 2 y-±-dx 2 is negative.) 



EI % =Pl ^ x) - 



■Pik—x) ; 



(1) 



whence, by taking the x anti-derivatives of both members 



M ^ Pilx -^- p ^ x - 



X 2 



)+c 



To find C, write out this equation for the point 0, where 
dy-i-dx=a and x=0, and we have C=EIoq', hence the 
equation for slope is 



FLEXURE ELASTIC CUETES. 257 

EI^=P(lx-^)-P l (l l x~^)+EI aa . (2) 

Again taking the x anti-derivatives, we have from (2) 

Ely =P ^-^-P^-J }+EIo«c+(O'=0) (3) 

(at Oboth x and?/ are —0 .*. (7'=0). In equations (1), (2), 
and (3) no valne of x is to be nsed <0 or >£, since for 
points in CB different relations apply, thus 

Equation of CB. — Fig. 227. Let the free body extend 
from B to a section made anywhere on O^.^moms.), as 
before, =0, gives 

EI % = - p ^- x) '• • • (4) 

(N.B. In (4), as in (1), EldPy-h-dx 2 is written equal to a neg- 
ative quantity because itself essentially negative ; for the 
curve is concave to the axis X in the first quadrant of the 
co-ordinate axes.) 

From (4) we have in the ordinary way (x-anti-deriv.) 

EI^=-P 1 (l^> —%)+0" . . (5/ 

ax A 

To determine C", consider that the curves CB and OC 
have the same slope (dy-r-dx) at G where x=l; hence put 
x—l in the right-hand members of (2) and of (5)' and 
equate the results. This gives G" = 1 / 4Pl 2 -\-EIa Q and .-. 



^nd.: Ely = ^x+Ela x-P l [l l ^-^l +C" . (6)' 
A A b 



258 MECHANICS OF ENGINEERING. 

At C, where x=l, both curves have the same ordinate; 
hence, by putting x=l in the right members of (3) and (6)' 
and equating results, we obtain C'"= — }4Pl 3 . .'. (6)' be 
comes 

EIy = y 2 PPx+EIa^x-P 1 \^-t^-^ . (6) 

as the Equation of CB, Fig. 227. But Oq is still an unknown 
constant, to find which write out (6) for the point B where 
x=l, and y=0, whence we obtain 

<*o= ^ n [Pl 3 -3PPl l +2Pm . . (7) 

a x = a similar form, putting P for P u and (^ — I) for I. 

235. Maximum Deflection in Case II. — Fig. 227. The or- 
dinate y m of the lowest point is thus found. Assuming 
l>/4li> it will occur in the curve OC. Hence put the 
dy-^dx of that curve, as expressed in equation (2), =0. 
Also for «q write its value from (7), having -put P 1 =Pl-^-l l , 
and we have 

P(l^-t )-P~ &«- %-)+ y 6 ^ (V-3U 1 +2l 1 ')= 

A Ly A l\ 

whence [x for max. 2/] = ^^^ — I) 

Now substitute this value of x in (3), also Oq from (7), and 
put Pi =Pl^-li, whence 

Max. Deflec.=2/max=V9 - -^- [Z 3 — 3^+2^] ^y^^—l). 

236. Case III. Horizontal Prismatic Beam Supported at Both 
Ends and Bearing a Uniformly Distributed Load along its Whole 
Length. — (The weight of the beam itself, if considered, 






FLEXURE. ELASTIC CURVES. 



259 



constitutes a load of this nature.) Let 1= the length 
of the beam and w— the weight, per unit of length, 
of the loading ; then the load coming upon any length x 
will be =wx 3 and the whole load =wl. By hypothesis w 
is constant. Fig. 228. From symmetry we know that the 




w««^ 



UmjlLLil, 



■w7 




Fig. 228. 

reactions at A and B are each =j4wl, that the middle of 
the neutral line is its lowest point, and the tangent line at 
is horizontal. Conceiving a section made at any point 
m of the neutral line at a distance x from 0, consider as 
free the portion of beam on the right of m. The forces 
holding this portion in equilibrium are yiwl, the reaction 
at B ; the elastic forces of the exposed surface at m, viz.: 
the tensions and compressions, forming a couple, and J 
the total she?r ; and a portion of the load, toQjjj—^x), The 
sum of the moments of these latter forces about the neu- 
tral axis of m, is the same as that of their resultant ; (i.e., 
their sum, since they are parallel), and this resultant acts in 
the middle of the length y 2 l — x. Hence the sum of these 
moments =w(*4l — x)j4{^l — x). Now putting 2 (mom. 
about neutral axis of m) =0 for this free body, we have 



*'& 



y2wi(y 2 ir-x)—y 2 w(y 2 ir--<c) 2 ; 



i.e., EI 



tfy_ 
dx 2 



wiyp—x 2 ) 



(1) 



260 



MECHANICS OF ENGINEERING. 



Taking the #-anti-derivative of both sides of (1), 

EI Tx = I /2w(ti l2 *- I A^)+(C=0) (2) 

as the equation of slope. (The constant is =0 since at 
both dy-^-dx and x are =0.) From (2), 



w 



^Iy=- 2 (}i^-%x^[C f =0] 



(3) 



which is the equation of the elastic curve ; throughout, 
i.e., it admits any value of x from x=-\-*4l to x= — y 2 l. 
This is an equation of the fourth degree, one degree high- 
er than those for the Curves of Cases I and II, where 
there were no distributed loads. If w were not constant, 
but proportional to the ordinates of an inclined right line, 
eq. (3) would be of the fifth degree ; if w were propor- 
tional to the vertical ordinates of a parabola with axis 
vertical, (3) would be of the sixth degree ; and so on. 

By putting x=j4l in (3) we have the deflection of be- 
low the horizontal through A and B, viz.: (with W= total 

load =wl) 

J 5 rf 5 Wf , 4) 



384 EI 384 



EI 



237. Case IV. Cantilevers. — A horizontal beam whose only 
support consists in one end being built in a wall, as in 
Fig. 229(a), or supported as in Fig. 
229(6) is sometimes called a canti- 
lever. Let the student prove that in 
Fig. 229(a) with a single end load P, 
the deflection of B below the tangent 
at Ois d= y PV '-T- El ~;the same state- 
ment applies to Fig. 229(6), but the 
tangent at is not horizontal if the 
beam was originally so. It can also 
be proved that the slope at B, Fig. 
229(a) (from the tangent at 0) is 







FLEXURE ELASTIC CURVES. 261 

PI 2 



Of 



2EI 



The greatest deflection of the elastic curve from the right 
line joining AB, in Fig. 229(6), is evidently given by the 
equation for y max. in § 235, by writing, instead of P of 
that equation, the reaction at in Fig. 229(6). This assumes 
that the max. deflection occurs between A and 0. If it 
occurs between and B put (l\—l) for I. 

If in Fig. 229(a) the loading is uniformly distributed 
along the beam at the rate of w pounds per linear unit, 
the student may also prove that the deflection of B below 
the tangent at is 

238. Case V. Horizontal Prismatic Beam Bearing Equal Ter- 
minal Loads and Supported Symmetrically at Two Points.— 
Fig. 231. Weight of beam neglected. In the preceding 
cases we have made use of the approximate form ElcPy-^dx 2 
in determining the forms of elastic curves. In the present 



^ncri 



pj 

—- 

Fig. 231. 




case the elastic curve from to C is more directly dealt 
with by employing the more exact expression UI— t o (see 
§ 231) for the moment of the stress-couple in any section. 
The reactions at and Care each =P, from symmetry. 
Considering free a portion of the beam extending from A 
to any section ra between and G (Fig. 232) we have, by 
putting 2 (mom. about neutral axis of ra)=0, 

p(i+x)- ^L—px^o .-. p= M- 
K p r pi 



262 



MECHANICS OF ENGINEERING. 



That is, the radius of curvature is the same at all points 
of 0(7; in other words OG is the arc of a circle with the 
above radius. The upward deflection of F from the right 
line joining and G can easily be computed from a knowl- 
edge of this fact. This is left to the student as also the 
value of the slope of the tangent line at (and G). The 
deflection of D from the tangent at C= l J B Pl s -r-EI ) as ip 
Fig. 229(a). 



SAFE LOADS IN FLEXURE, 



239. Maximum Moment. — As we examine the different sec- 
tions of a given beam under a given loading we find differ- 
ent values of p, the normal stress per unit of area in the 
outer element, as obtained from eq. (5) § 229, viz.: 



pi _ 



e 



(i) 



in which J is the " Moment of Inertia " (§ 85) of the plane 
figure formed by the section, about its neutral axis, e the 
distance of the most distant (or outer) fibre from the neu- 
tral axis, and M the sum of the moments, about this neu- 
tral axis, of all the forces acting on the free body of which 
the section in question is one end, exclusive of the stresses 
on the exposed surface of that section. In other words 
M is the sum of the moments of the forces which balance 
the stresses of the section, these moments being taken 
about the neutral axis of the section under examination. 
For the prismatic beams of this chapter e and /are the 
same at all sections, hence p varies with M and becomes a 
maximum when M is a maximum. In any given case the 
location of the " dangerous section" or section of maximum 
M, and the amount of that maximum value may be deter- 
mined by inspection and trial, this being the only method 
(except by graphics) if the external forces are detached. 



ELEXTTKE SAFE LOADS. 263 

If, however, trie loading is continuous according to a de- 
finite algebraic law the calculus may often be applied, 
taking care to treat separately each portion of the beam 
between two consecutive reactions of supports, or detached 
loads. 

As a graphical representation of the values of M along 
the beam in any given case, these values may be conceived 
laid off as vertical ordinates (according to some definite 
scale, e.g. so many inch -lbs. of moment to the linear inch 
of paper) from a horizontal axis just below the beam. If 
the upper fibres are in compression in any portion of the 
beam, so that that portion is convex downwards, these or- 
dinates will be laid off below the axis, and vice versa ; for 
it is evident that at a section where M= 0, p also =0, i.e., 
the character of the normal stress in the outermost fibre 
changes (from tension to compression, or vice versa) when 
M changes sign. It is also evident from eq. (6) § 231 that 
the radius of curvature changes sign, and consequently the 
curvature is reversed, when M changes sign. These mo- 
ment ordinates form a Moment Diagram, and the extremities 
a Moment Curve. 

The maximum moment, M m9 being found, in terms of 
the loads and reactions, we must make the p of the " dan- 
gerous section," where M= M m9 equal to a safe value B' 9 
and thus may write 

— = Ma ...» (2) 



Eq. (2) is available for finding any one unknown quanti- 
ty, whether it be a load, span, or some one dimension of 
the beam, and is concerned only with the Strength, and not 
with the stiffness of the beam. If it is satisfied in any 
given case, the normal stress on all elements in all sections 
is known to be = or <i?', and the design is therefore safe 
in that one respect. 

As to danger arising from the shearing stresses in any 



264 



MECHANICS OF ENGLtfEEKDsG. 



section, the consideration of the latter will be taken up in 
a subsequent chapter and will be found to be necessary 
only in beams composed of a thin web uniting two flanges. 
The total shear, however, denoted by J t bears to the mo- 
ment M, an important relation of great service in deter- 
mining M m . This relation, therefore, is presented in the 
next article. 

240, The Shear is the First x-Derivative of the Moment. — 
Fig. 233. (x is the distance of any section, measured parallel 
\/ to the beam from an arbitrary 

pdF origin). Consider as free a ver- 
tical slice of the beam included 
between any two consecutive 
vertical sections whose distance 
apart is dx. The forces acting 
are the elastic forces of the two 
internal surfaces now laid bare, 
and, possibly, a portion, wdx t 
of the loading, which at this 
part of the beam has some intensity =w lbs. per running 
linear unit. Putting I'fmom. about axis JV')=0 we have 
(noting that since the tensions and compressions of section 
N form a couple, the sum of their moments about N' is 
just the same as about N,)' 

e e 2 



pdF 



IpdF - 




vi- 



P'l _ 



But r±=M, the Moment of the left hand section/il =M>, 

e e 

that of the right ; whence we may write, after dividing 
through by dx and transposing, 



M'—M 
dx 



J-\-w 



dx 



;. dM t 
dx 



(3) 



for w ^ vanishes when added to the finite J, and M' — M= 
dM= increment of the moment corresponding to the incre- 
ment, dx, of x. This proves the theorem. 






FliEXUEE.SAFB LOADS. 265 

Now the value of x which renders M a maximum or 
minimum would be obtained by putting the derivative 
dM- : rdx= zero ; hence we may state as a 

Corollary. —At sections where the moment is a maximum 
or minimum the shear is zero. 

The shear J at any section is easily determined by con- 
sidering free the portion of beam from the section to either 
end of the beam and putting ^(vertical components) =0. 

In this article the words maximum and minimum are 
used in the same sense as in calculus ; i.e., graphically, 
they are the ordinates of the moment curve at points 
where the tangent line is horizontal. If the moment curve be 
reduced to a straight line, or a series of straight lines, it 
has no maximum or minimum in the strict sense just 
stated ; nevertheless the relation is still practically borne 
out by the fact that at the sections of greatest and least 
ordinates in the moment diagram the shear changes sign 
suddenly. This is best shown by drawing a shear diagram, 
whose ordinates are laid off vertically from a horizontal 
axis and under the respective sections of the beam. They 
will be laid off upward or downward according as J" is 
found to be upward or downward, when the free body con- 
sidered extends from the section toward the right. 

In these diagrams the moment ordinates are set off on 
an arbitrary scale of so many inch -pounds, or foot-pounds, 
to the linear inch of paper ; the shears being simply 
pounds, or some other unit oi force, on a scale of so many 
pounds to the inch of paper. The scale on which the 
beam is drawn is so many feet, or inches, to the inch of 
paper. 

241. Safe Load at the Middle of a Prismatic Beam Support- 
ed at the Ends. — Fig. 234. The reaction at each support 
is j^P. Make a section n at any distance x<J- from B. 
Consider the portion nB free, putting in the proper elas- 
tic and external forces. The weight of beam is neglected. 
From I(mom. about n)=0 we have 



266 



MECHANICS OF ENGINEERING. 



^=^x; Le.,M=y 2 Px 

e 2 

Evidently M is proportional to x, and the ordinates repre- 
senting it will therefore be limited by the straight line 




,Fig. 234. 



B'R, forming a triangle B'BA'. From symmetry, another 
triangle O'RA' forms the other half of the moment dia- 
gram. From inspection, the maximum M is seen to be in 
the middle where x= y£l t and hence 



(ifmax.)=i!f ra =^PZ 



(1) 



Again by putting JYvert. compons.)=0, for the free body 
nB we have 

H 

and must point downward since ~ points upward. Hence 
the shear is constant and =^Pat any section in the right 
hand half. If n be taken in the left half we would have, 
nB being free, from -(vert. com.)=0, 



FLEXURE. SAFE LOADS. 267 

the same numerical value as before ; but J* must point up- 
ward, since | at B and J at n must balance the downward 
P at A, At A, then, the shear changes sign suddenly, 
that is, passes through the value zero; also at A, M is a 
maximum, thus illustrating the statement in § 240. Notice 
the shear diagram in Fig. 234. 

To find the safe load in this case we write the maximum 
value of the normal stress, p,= B l , a safe value, (see table 
in a subsequent article) and solve the equation for P. 
But the maximum value of p is in the outer fibre at A, 
since if for that section is a maximum. Hence 

~ = %n (2) 

is the equation for safe loading in this case, so far as the 
normal stresses in any section are concerned. 

Example. — If the beam is of wood and has a rectangu- 
lar section with width b= 2 in., height fa*= 4 in., while its 
length 1= 10 ft., required the safe load, if the greatest nor- 
mal stress is limited to 1,000 lbs. per sq. in. Use the 
pound and inch. From § 90 I=% b¥=%X 2x64= 10.66 
biquad. inches, while e=l=2 in. 

.-. P= 4^x1,000x10,66 „ im lbg> 

le 120x2 

242. Safe Load Uniformly Distributed along a Prismatic Beam 
Supported at the Ends.— Let the load per lineal unit of the 
length of beam be =w (this can be made to include the 
weight of the beam itself). Fig. 235. From symmetry, 

each reaction == y 2 wl. For the free body nO we have, put* 
ting Ifmom. about n)=0, -.- -+i 

Pi Wl / \ X tit W /7 ,v 



268 



MECHANICS OF ENGINEERING. 



which gives Mfor any section by making x vary from 
to I. Notice that in this case the law of loading is con- 
tinuous along the whole length, and that hence the mo- 
ment curve is continuous for the whole length. 



W=m/Z 




Fig. 235. 



To find the shear J, at n, we may either put ^(vert. com- 
pons.)= for the free body, whence J= y 2 wl — wx 9 and must 
therefore be downward for a small value of x ; or, employ- 
ing § 240, we may write out dM-r-dx f which gives 



dM_^ {l _ %x) 



dx 



(i) 



the same as before. To find the max. M 9 or M mi put J=0, 
which gives x=y 2 l. This indicates a maximum, for when 
substituted in d 2 M-r-dx 2 , i.e., in — w, a negative result is 
obtained. Hence M m occurs at the middle of the beam and 
its value is 



M m = %wl 2 ; 



EI 



y w l 2 =%Wl 



(2) 



the equation of safe loading. W= total load=wZ. 

It can easily be shown that the moment curve is a por- 



FLEXUKE.SAFE LOADS. 



26$ 



jion of a parabola, whose vertex is at A" under the mid- 
Jib of the beam, and axis vertical. The shear diagram 
consists of ordinates to a single straight line inclined to 
its axis and crossing it, i.e., giving a zero shear, under the 
middle of the beam, where we find the max. M. 

If a frictionless dove-tail joint with vertical faces were 
introduced at any locality in the beam and thus divided 
the beam into two parts, the presence of J would be made 
manifest by the downward slipping of the left hand part 
on the right hand part if the joint were on the right of the 
middle, and vice versa, if it were on the left of the middle. 
This shows why the ordinates in the two halves of the 
shear diagram have opposite signs. The greatest shear 
is close to either support and is J m =j4wL 

243. Prismatic Beam Supported at its Extremities and Loaded 
in any Manner. Equation for Safe Loading. — Fig. 236. Given 

the loads P l9 P 2 , and P 3 , whose 
distances from the right sup- 
port are l lf l 2 , and 1$ ; ,required 
the equation for safe loading ; 
i.e., find M m and write it = 
R'l+e. 

If the moment curve 
continuous, i.e., if M 



r r r 



W-pcg- 




were 



were a 



continuous function of x from 
end to end of the beam, we 
could easily find M m by making 
fig. 236. dM- : rdx=0, i.e., e/=0, and sub- 

stitute the resulting value of x in the expression for M. 
But in the present case of detached loads, J is not zero, 
necessarily, at any section of the beam. Still there is 
some one section where it changes sign, i.e., passes sud- 
denly through the value zero, and this will be the section 
of greatest moment (though not a maximum in the strict 
sense used in calculus). By considering any portion n O 
as free, J is found equal to the Reaction at Diminished "by 
the Loads Occuring Between n and 0. The reaction at B is 



270 MECHANICS OF ENGINEERING. 

obtained by treating the whole beam as free (in which case 
no elastic forces come into play) and putting Ifmom. 
about O)=0; while that at O,=P =P 1 +P 2 +P s —P B 
If n is taken anywhere between and E, J=P 

E " F y J=P -P 1 
F « H, J=P -P 1 -P 2 
H " B, J^P () -P l -P 2 -P z 
This last yalue of J" also == the reaction at the other 
support, B. Accordingly, the shear diagram is seen to 
consist of a number of horizontal steps. The relation 
J=dM- L rdx is such that the slope of the moment curve is 
proportional to the ordinate of the shear diagram, and 
that for a sudden change in the slope of the moment curve 
there is a sudden change in the shear ordinate. Hence in 
the present instance, J being constant between any two 
consecutive loads, the moment curve reduces to a straight 
line between the same loads, this line having a different 
inclination under each of the portions into which the beam 
is divided by the loads. Under each load the slope of the 
moment curve and the ordinate of the shear diagram change 
suddenly. In Fig. 236 the shear passes through the value 
zero, i.e., changes sign, at F; or algebraically we are sup- 
posed to find that P — P x is + while P — P x — P 2 is — , in 
the present case. Considering FO, then, as free, we find 
M m to be 

M m = P l 2 — Pi(l 2 —li) and the equation for safe loading is 

EI^p^-p^-i,) (i) 

e 

(i.e., if the max. ill" is at F). It is also evident that the 
greatest shear is equal to the reaction at one or the other 
support, whichever is the greater, and that the moment 
at either support is zero. 

The student should not confuse the moment curve, which 



FLEXUKE. SAFE LOADS. 



271 



is entirely imaginary, with the neutral line (or elastic 
curve) of the beam itself. The greatest moment is not 
necessarily at the section of maximum deflection of the 
neutral line (or elastic curve). 

For the case in Fig. 236 we may therefore state that the 
max. moment, and consequently the greatest tension or 
compression in the outer fibre, will be found in the sec- 
tion under that load for which the sum of the loads (in- 
cluding this load itself) between it and either support first 
equals or exceeds the reaction of that support. The 
amount of this moment is then obtained by treating as free 
either of the two portions of the beam into which this 
section divides the beam. 

244. Numerical Example of the Preceding Article. — Fig. 237. 
Given P lt P 2 , P B , equal to ]/ 2 ton, 1 ton, and 4 tons, re- 




spectively ; \ =5 feet, l 2 = 7 feet, and l s = 10 feet ; while the 
total length is 15 feet. The beam is of timber, of rectan- 
gular cross-section, the horizontal width being 5=10 
inches, and the value of R' (greatest safe normal stress), 
= y 2 ton per sq. inch, or 1,000 lbs. per sq inch. 



272 MECHANICS OF ENGINEERING. 

Required the proper depth h lor the beam, for safe load- 
ing. 

Solution. — Adopting a definite system of units, viz., the 
inch-ton-second system, we must reduce all distances such 
as I, etc., to inches, express all forces in tons, write ~R' = }£ 
(tons per sq. inch), and interpret all results by the same sys- 
tem. Moments will be in inch-tons, and shears in tons. 
[N. B. In problems involving the strength of materials 
the inch is more convenient as a linear unit than the foot, 
since any stress expressed in lbs., or tons, per sq. inch, is 
numerically 144 times as small as if referred to the square 
foot.] 

Making the whole beam free, we have from moms, about 
O, P B =± [}4x 60+1x84+4x120] =3.3 tons.-. P =5.5— 
3.3=2.2 tons. 

The shear anywhere between and F is J= P =2.2 tons. 

F and F is J =2.2— J^ =1.7 

tons. 
The shear anywhere between F and His J =2.2 — y 2 — 1 = 

0.7 tons. 
The shear anywhere between B. and B is J = 2.2 — y 2 — 1 

—4 =—3.3 tons. 

Since the shear changes sign on passing H, .-. the max.. 

moment is at H; whence making HO free, we have 

M at H=M m =2.2 x 120—^ x 60—1 x 36 =198 inch -tons. 

/?' T 
For safety M m must =- — , in which R = }& ton per sq. 

e 

inch, e = ^h = i^ of unknown depth of beam, and 7, §90, = 

A bh z , with 6=10 inches 

!li. ^.f-xl0/i 3 =198; or ¥ =237.6.-. £=15.4 inches. 

245. Comparative Strength of Rectangular Beams. — For such 
a beam, under a given loading, the equation for safe load- 
ing is 

^l=M m i. e. % R lV=M m .... (1) 

Q 



FLEXUKE. SAEE LOADS. 273 

whence the following is evident, (since for the same length, 
mode of support, and distribution of load, M m is propoT- 
tional to the safe loading.) 

For rectangular prismatic beams of the same length, 
same material, same mode of support and same arrange- 
ment of load : 

(1) The safe load is proportional to the width of beams 
having the same depth (A). 

(2) The safe load is proportional to the square of the 
depth of beams having the same width (6). 

(3) The safe load is proportional to the depth of beams 
having the same volume (i. e. the^same bh), 

(It is understood that the sides of the section are hori- 
zontal and vertical respectively and that the material is 
homogeneous.) ' 

246. Comparative Stiffness of Rectangular Beams.— Taking the 
deflection under the same loading as an inverse measure 
of the stiffness, and noting that in §§ 233, 235, and 236, 
this deflection is inversely proportional to I=±bh s = 
the "moment of inertia" of the section about its neutral 
axis, we may state that : 

For rectangular prismatic beams of the same length, 
same material, same mode of support, and same_ loqjiing : 

(1) The stiffness is proportional to the width for beams 
of the same depth. P .* --.. 

(2) The stiffness is proportional to the cube- o| the 
height for beams of the same width (b). 

(3) The stiffness is proportional to the square of the 
depth for beams of equal volume (bhl). , v „* • 

(4) If the length alone vary, the ^stiffness i& inversely 
proportional to the cube of the length. 

247. Table of Moments of Inertia. — These are here recapitu- 
lated for the simpler cases, and also the values of e. the 
distance of the outermost fibre Irom the axis. •>.. 

Since the stiffness varies as /(other things being -equal), 



274 



MECHANICS OF ENGINEERING. 



while the strength varies as I-r- e, it is evident that a 
square beam has the same stiffness in any position (§89), 
while its strength is greatest with one side horizontal, for 
then e is smallest, being ^}£b. 

Since for any cross-section 1= CdF z\ in which 3=the 

distance of any element, dF, of area from the neutral axis, 
a beam is made both stiffer and stronger by throwing 
most of its material into two flanges united by a vertical 
web, thus forming a so-called " I-beam " of an I shape. But 
not without limit, for the web must be thick enough to 
cause the flanges to act together as a solid of continuous 
substance, and, if too high, is liable to buckle sideways, 
thus requiring lateral stiffening. These points will be 
treated later. 



»1 




1 

i 

i 


' 


FFh 






\ 














(a) 








SECTION. 



Bectaogjet width = b, depth = h (vertical) 

Hollow "Rectangle, symmet. about neutral axis. See \ 
Fig. 238 (a) J 

Trianglev width =5, height = h, neutral axis parallel I 
%o base (horizontal). S 

Circle of rtdius r 

Ring of concentri® circles. Fig. 938 (b) 

Ehombust Ftg. 236 (c) h = diagonal which ts vertical. 

Square with Bide b vertical. 

" ** b at 45° with horiz, 



Via bh s 
Via [6i Ai»-& a h\l 

V48 ^ S 

Via * 4 
Via & 4 



e 



r 

%b 



248. Moment of Inertia of I-beams, Box-girders, Etc. — In 

common with other large companies, the N. J. Steel and 



FLEXURE. SAFE LOADS. 



275 



Iron Co. of Trenton, N. J. (Cooper, Hewitt & Co.) manu- 
facture prismatic rolled beams of wrought-iron variously 
called /-beams, deck -beams, rails, and " shape iron," (in- 
cluding channels, angles, tees, etc., according to the form 
of section.) See fig. 239 for these forms. The company 



^7^ 



d- 



^» 



ii 



T 



fcBEAM.. CHANNEL. DECK-BEAM; RAIL. 

Fia. 239. 



TEE. 



publishes a pocket-book giving tables of quantities rela- 
ting to the strength and stiffness of beams, such as the 
safe loads for various spans, moments of inertia of their 
sections in various positions, etc., etc. The moments of 
inertia of /-beams and deck -beams are computed accord- 
ing to §§ 92 and 93, with the inch as linear unit. The 
/-beams range from 4 in. to 20 inches deep, the deck- 
beams being about 7 and 8 in- deep. 

For beams of still greater stiffness and strength com- 
binations of plates, channels, angles, etc., are riveted to- 
gether, forming " built-beams," or " plate girders." The 
proper design for the riveting of such beams will be ex- 
amined later. For the present the parts are assumed to 
act together as a continuous mass. For example, Fig. 240 
shows a " box -girder," formed of two " channels " and 
two plates riveted together. If the axis of symmetry, N, 

h — — ?,, 4 i s to be horizontal it becomes the neu- 

axis. Let 0*= the moment of iner- 
of one channel (as given in the 
pocket-book mentioned) about the axis 
N perpendicular to the web of the chan- 
nel. Then the total moment of inertia of 
the combination is (nearly) 



^ 



SP 



p[P*tral 

tia 



33 



Fig. 240. 



J N =2(7+25^ 2 — 4d'f (d— y 2 if 



(1) 



276 



MECHANICS OF ENGINEERING. 



In (1), b, t, and d are the distances given in Fig. 240 (d ex- 
tends to the middle of plate) while d' and t' are the length 
and width of a rivet, the former from head to head 
(i.e., d' and t' are the dimensions of a rivet-hole). 

For example, a box-girder of wrought-iron is formed of 
two 15-inch channels and two plates 10 inches wide and 1 
inch thick, the rivet holes ^ in. wide and 1^ in. long. 
That is, 6=10; t=l; d=8; t' = ^\ and d'=l^ inches. 
Also from the pocket-book we find that for the channel in 
question, (7=376 biquadratic inches. Hence, eq. (1) 



4 = 752+2x10x1x64—4x1x^(8- 



%y- 



1737biquadr.in. 



Also, since in this instance e = 8)4 inches, and 12000 
lbs. per sq. inch (or 6 tons per sq. in.) is the value for B' 
(= greatest safe normal stress en the outer element of any 
cross-section) used by the Trenton Co. (for wrought iron), 

BI_ 12000x1 737 

e ~ 8.5 



we have 



=2451700 inch-lbs. 



That is, the box-girder can safely bear a maximum mo- 
ment, M m9 = 2451700 inch-lbs. = 1225.8 inch-tons, as far 
as the normal stresses in any section are concerned. 
(Proper provision for the shearing stresses in the section, 
and in the rivets, will.be considered later). 



249. Strength of Cantilevers.- 



d=±.s 



In Fig. 241 with a single 
concentrated load P at the 
projecting extremity, we 
3 ° easily find the moment at 
n to be M =Px, and the 
/ max. moment to occur at 
the section next the wall, 
/; its value being M m =Pl. 
The shear, J, is constant, 
fig. 242. an( j = p at all sections. 

The moment and shear diagrams are drawn in accordance 
with these results. 



• j 
B 


w Ha?-*] 


! 1 


b: 


I] 1 






Fig. 241. 



FLEXURE. SAFE LOADS. 



If the load W = id is uniformly distributed on the can- 
tilever, as in fig. 242, by making nO free we have, putting 
2(mom. about n) = 0, 



P i_ 

e 



■icx 



2.*.M=y 2 wtf.:M m =y 2 wi 2 = y 2 



Wl. 



Hence the moment curve is a parabola, whose vertex is at 
0' and axis vertical. Putting I (vert, compons.) = we 
obtain J = wx. Hence the shear diagram is a triangle, 
and the max. J= wl = W, 



250. Resume" of the Four Simple Cases. — The following table 
shows the values of the deflections under an arbitrary 
load P, or W, (within elastic limit), and of the safe load ; 





Cantilevers . 


Beams with tv\ 


o end supports. 




With one end 
loadP 
Fig. 241 


With unif . load 
W=wl 
Fig. 242 


Load Pin 
middle 
Fig. 234 


Unif. loaa 
W=wl 
Fig. 235 


Deflection 

( Safe load (from ?— 

{ e 

( = Mm) 


E'l 

le 


EI 

■ E'l 

2 -77 


1 El 3 

48' EI 

4 E'I 

le 


5 WP 

384* EI 

le 


Relative strength 


1 


2 


4 


8 
128 


j Relative stiffness 
| under same load 


1 


8 / 3 


16 


16 
5 


1 Relative stiffness 
I under safe load 


1 


4 / 3 


4 


j Max. shear = Jin, (and 
( location, 


P, (at wall) 


W, (at wall) 


HP, (at supp). 


y 2 W, (at supp) t 



also the relative strength, the relative stiffness (under the 
same load), and the relative stiffness under the safe load, 
for the same beam. 

The max. shear will be used to determine the proper 
web-thickness for I-beams and " built-girders." The stu- 
dent should carefully study the foregoing table, noting 
especially the relative strength, stiffness, and stiffness 
under safe load, of the same beam. 

Thus, a beam with two end supports will bear a double 



278 MECHANICS OF ENGINEERING. 

load, if uniformly distributed instead of concentrated in 
the middle, but will deflect J^ more ; whereas with a given 
load uniformly distributed the deflection would be only 
$/% of that caused by the same load in the middle, provided 
+ he elastic limit is not surpassed ii> either case. 

251. R', etc. For Various Materials.— The formula^/= 3I mi 

e 
from which in any given case of flexure we can compute 
the value of p m , the greatest normal stress in any outer 
element, provided all the other quantities are known, 
holds good theoretically within the elastic limit only. 
Still, some experimenters have used this formula for the 
rupture of beams by flexure, calling the value of p m thus 
obtained the Modulus of Rupture, R. R may be found to 
differ considerably from both the T or G of § 203 with 
some materials and forms, being frequently much larger. 
This might be expected, since even supposing the relative 
extension or compression (i.e., strain) of the fibres to be 
proportional to their distances from the neutral axis as 
the load increases toward rupture, the corresponding 
stresses, not being proportional to these strains , beyond the 
elastic limit, no longer vary directly as the distances from the 
neutral axis ; and the neutral axis does not pass through the 
centre of gravity of the section, necessarily. 

The following table gives average values for R, R', R"> 
and E for the ordinary materials of construction. E, the 
modulus of elasticity for use in the formulae for deflection, 
is given as computed from experiments in flexure, and is 
nearly the ?ame as E t and E c . 

In any example involving R', e is usually written equal 
to the distance of the outer fibre from the neutral axis, 
whether that fibre is to be in tension or compression ; 
since in most materials not only is the tensile equal to the 
compressive stress for a given strain (relative extension 
or contraction) but the elastic limit is reached at about 
the same strain both in tension and compression. 



FLEXURE. SAFE LOADS. 279 

Table foe Use in Examples in Flexure. 





Timber. 


Cast Iron. 


Wro't Iron. 


Steel. 


Max. safe stress in outer fi- ) 
bre =-B'<Ibs. per sq. inch). J 


1,000 


6,000 in tens. 
12,000 in comp. 


12,000 


15.000 

to 
40,000 


Stress in outer fibre at Elas. ) 






17.000* 
to 


30.000 
and upward. 


limit =J!''(lbs. per sq. in.) ) 






35,000 


" Modul. of Rupture " ) 
=i?=lbs. per sq. inch. ) 


4,000 

to 

20,000 


40,000 


50,000 


120,000 
Hard Steel. 


A'=Mod. of Elasticity, | 
=lbs. per sq. inch. ) 


1,000,000 

to 
3,000,000 


17,000,000 


25,003,000 


30,000,000 



3 



In the case of cast iron, however, (see § 203) the elastic 
limit is reached in tension with a stress =9,000 lbs. per 
sq. inch and a relative extension of ijL of one per cent., 
while in compression the stress must be about double to 
reach the elastic limit, the relative change of form (strain) 
being also double. Hence with cast iron beams, once 
largely used but now almost entirely displaced by rolled 
wrought iron beams, an economy of material was effected 
by making the outer fibre on the compressed side twice 
as far from the neutral axis as that on the stretched side. 
Thus, Fig. 243, cross-sections with unequal flanges were 
used, so proportioned that the centre of 
gravity was twice as near to the outer 
fibre in tension as to that in compression, 
i.e,, e 2 = 2e [ ; in other words more material 
is placed in tension than in compression. 
The fibre A being in tension (within elas- 
tic limit), that at B, since it is twice as far from the neu- 
tral axis and on the other side, is contracted twice as much 
as A is extended ; i.e., is under a compressive strain 
double the tensile strain at A, but in accordance with the 
above figures its state of stress is proportionally as much 
within the elastic limit as that of A. 

Steel beams are gradually coming into use, and may ul- 
timately replace those of wrought iron. 

* In the tests by U. S. Gov. in 1879 with I-beams, R" ranged from 25,000 
to 38,000, and the elastic limit was reached with less stress in the large 
than in the smaller beams. Also, for the same beam, R" decreased with 
larger spans. 



Fig. 243. 



28ft MECHANICS -OF ENGINEERING. 

The great range of values of R for timber is due not 
only to the fact that the various kinds of wood differ 
widely in strength, while the behavior of specimens of 
any one kind depends somewhat on age, seasoning, etc., 
but also to the circumstance that the size of the beam un- 
der experiment has much to do with the result. The ex- 
periments of Prof. Lanza at the Mass. Institute of Tech- 
nology in 1881 were made on full size lumber (spruce), of 
dimensions such as are usually taken for floor beams in 
buildings, and gave much smaller values of R (from 3,200 
to 8,700 lbs. per sq. inch) than had previously been ob- 
tained. The loading employed was in most cases a con- 
centrated load midway between the two supports. 

These low values are probably due to the fact that in 
large specimens of ordinary lumber the continuity of it& 
substance is more or less broken by cracks, knots, etc., 
the higher values of most other experimenters having 
been obtained with small, straight-grained, selected pieces, 
from one foot to six feet in length. 

The value i2' =12,000 lbs. per sq. inch* is employed by 
the N. J. Iron and Steel Co. in computing the safe loads 
for their rolled wrought iron beams, with the stipulation 
that the beams (which are high and of narrow width) must 
be secure against yielding sideways. If such is not the 
case the ratio of the actual safe load to that computed with 
R!= 12,000 is taken less and less as the span increases. 
The lateral security referred to may be furnished by the 
brick arch-filling of a fire-proof floor, or by light lateral 
bracing with the other beams. 

252. Numerical Examples, — Example 1. — A square bar of 
wrought iron, \y 2 in. in thickness is bent into a circular 
arc whose radius is 200 ft., the plane of bending being par- 
allel to the side of the square. Eequired the greatest nor- 
mal stress p m in any outer fibre. 

Solution. From §§ 230 and 231 we may write 

_ =£— .'. p=eU-r-p, i.e., is constant. 

P e 

* For their soft-steel beams this company uses 16,000 lbs. per sq. inch. 



FLEXURE. SAFE LOADS. 281 

For the units inch and pound (viz. those of the table in § 
251) we have e=% in., p =2,400 in., and ^=25,000,000 lbs. 
per sq. inch, and .*. 

p=j9 m =^x 25,000,000-i-2,400 =7,812 lbs. per sq. in., 

which is quite safe. At a distance of y 2 inch from the 
neutral axis, the normal stress is =[}4-r-}{]pm = %Pm = 
5,208 lbs. per sq. in. (If the force-plane (i.e., plane of 
bending) were parallel to the diagonal of the square, e 
would =^X 1.5^2 inches, giving p m = [7,812x^/2 ] lbs. 
per sq. in.) § 238 shows an instance where a portion, OC, 
Fig. 231, is bent in a circular arc. 

Example 2. — A hollow cylindrical cast-iron pipe of radii 
3 y 2 and 4 inches * is supported at its ends and loaded in 
middle (see Fig. 234). Ee quired the safe load, neglecting 
the weight of the pipe. From the table in § 250 we have 
for safety 

P=4 *L 

le 

From § 251 we put i?' = 6,000 lbs. per sq. in.; and from § 
247 I— ^(r^ — r 2 4 ); and with these values, r 2 being = 1, r\= 
4, e=r r =4, ~=-y- and Z=144 inches (the inch must be the 
unit of length since B' = 6,000 lbs. per sq. inch) we have 

P=4x 6,000x^. -f-(256-150)-r-[144x4] .\ P= 3,470 lbs. 
The weight of the beam itself is G= Vr, (§ 7), i.e., 

^=-( n 2 -r 2 2 )? r = ^(16-12^)144x^=443 lbs. 

(Notice that y, here, must be lbs., per cubic inch). This 
weight being a uniformly distributed load is equivalent to 
half as much, 221 lbs., applied in the middle, as far as the 
strength of the beam is concerned (see § 250), .*. P must be 
taken =3,249 lbs. when the weight of the beam is consid- 
ered. 

* And 12 feet in length between supports. 



15 



282 MECHANICS OF ENGINEERING. 

Example 3. — A wrought-iron rolled I-beam supported 

at the ends is to be loaded uniformly Fig. 235, the span 

being equal to 20 feet. Its cross-section, Fig. 244, has a 

r— 5 ^ depth parallel to the web of 15 

^r"^ inches, a flange width of 5 inches. 

[\ f\ In the pocket book of the Trenton 

_t/ __j_^ ^ ^ ^ ca lled a 15-inch light I- 

beam, weighing 150 lbs. per yard, 
fig. 244. with a moment of inertia =523. bi- 

quad. inches about a gravity axis perpendicular to the 
web (i.e., when the web is vertical, the strongest position) 
and = 15 biq. in. about a gravity axis parallel to the web 
(i.e., when the web is placed horizontally). 

First placing the web vertically, we have from § 250,, 

W,= Safe load, distributed, =8^5. With £'=12,000, 

le x 

i!=523, 1=240 inches, e l =7}4 inches, this gives 
W,= [8x12,000x523] -H[240xf ] =27,893 lbs. 

But this includes the weight of the beam, G—20 ft. X^-lbs. 
= 1,000 lbs.; hence a distributed load of 26,902 lbs., or 13.45 
tons may be placed on the beam (secured against lateral 
yielding). (The pocket-book referred to gives 13.27 tons 
as the safe load.) 

Secondly, placing the web horizontal, 

^ 2 =8 4-^ = [8xl2,000xl5]^[240x 5 / 2 ]-^ of W x 

or only about 1 / 12 of W Y . 

Example 4. — Kequired the deflection in the first case of 
Ex. 3. From § 250 the deflection at middle is 



d r 



5 


WJ? 


5 m'i x i 3 _ 5 


R' 


P 


384 


EI, 


384' le { ' EI X 48 


' E 


€j 



ELEXUKE. SAFE LOADS. 283 



, 5 12,000 (240) 2 r , , n 

i.e., d 1 =—- . — ' . v ; ; (men and pound) 



48 25,000,000 



.-. ^=0.384 in. 
Example 5. — A rectangular beam of yellow pine, of width 
6=4 inches, is 20 ft. locg, rests on two end supports, and is 
to carry a load of 1,200 lbs. at the middle ; required the 
proper depth h. From § 250 

le I 12 " }A 

.-. h 2 =6Pl- : r4:Il'b. For variety, use the inch and fom. For 
this system of units P=0.60 tons, i?'=0.50 tons per sq. in., 
Z=240 inches and b= 4 inches. 

.-. Zi 2 =(6x0.6x240)4-(4x0.5x4)=108sq. in. ,\ h =10.4 in. 

Example 6. — Suppose the depth in Ex. 5 to be deter- 
mined by the condition that the deflection shall be = 1 / s[a 
of the span or length. We should then have from § 250 

d= 1 1=1 Ht 

500 48 EI 

Using the inch and ton, with i?= 1,200,000 lbs. per sq. in., 
which = 600 tons per sq. inch, and 7= 1 / 12 6A 3 , we have 

U= 500x0.60x240x240x12 =L800 ••• ^=12-2 m. 

48x600x4 

As this is > 10.4 the load would be safe, as well. 

Example 7. — Required the length of a wro't iron pipe 

supported at its extremities, its internal radius being 2^ 

in., the external 2.50 in.> that the deflection under its own 

weight may equal 1 / m of the length. 579.6 in. Ans. 

Example 8. — Fig. 245. The wall is 6 feet high and one 

foot thick, of common brick work 

/_- i ' i ' i ' i ' I (see § 7) and is to be borne by an 

/-beam in whose outer fibres no 

greater normal stress than 8,000 

Ej~ E lbs. per sq. inch is allowable. If 

fig. 345. a number of I-beams is available,, 



284 MECHANICS OF ENGINEERING. 

ranging in height from 6 in. to 15 in. (by whole inches), 
which one shall be chosen in the present instance, if their 
cross-seetions are Similar Figures, the moment of inertia of 
the 15-inch beam being 800 biquad. inches ? 

The 12-inch beam. Ans. 



SHEARING STRESSES IN FLEXURE. 

253. Shearing Stresses in Surfaces Parallel to the Neutral 
Surface, — If a pile of boards (see Fig. 246) is used to sup- 
port a load, the boards being free to slip on each other, it 
is noticeable that the ends overlap, although the boards 



i 1 




Fig. 246. Fig. 247. 

-are of equal length (now see Fig. 247) ; i.e., slipping has 
occurred along the surfaces of contact, the combina- 
tion being no stronger than the same boards side by- 
side. If, however, they are glued together, piled as in the 
former figure, the slipping is prevented and the deflection 
is much less under the same load P. That is, the com- 
pound beam is both stronger and stiffer than the pile of 
loose boards, but the tendency to slip still exists and is 
known as the " shearing stress in surfaces parallel to the 
neutral surface." Its intensity per unit of area will now 
be determined by the usual " free-body " method. In Fig. 
248 let AN' be a portion, considered free, on the left of any 




SHEAR IX FLEXURE. 



285 



section N', of a prismatic beam slightly bent under forces 
in one plane and perpendicular to the beam. The moment 
equation, about the neutral axis at N\ gives 



*—=M'; whence p'= — =- 
e I 



(i) 



Similarly, with AN as a free body, NN' being =dx, 

t—=3f; whence p=^ r . 
e I 



(2) 



p and p' are the respective normal stresses in the outer 
fibre in the transverse sections N and N' respectively. 

Now separate the block NN', lying between these two 
consecutive sections, as a free body (in Fig. 249). And 




PART OF J 




8ABT0F 



furthermore remove a portion of the top of the latter block, 
the portion lying above a plane passed parallel to the neu- 
tral surface and at any distance z" from that surface. This 
latter free body is shown in Fig. 250, with the system of 
forces representing the actions upon it of the portions taken 
away. The under surface, just laid bare, is a portion of a sur- 
face (parallel to the neutral surface) in which the above men- 
tioned slipping, or shearing, tendency exists. The lower por- 
tion (of the block NN') which is now removed exerted this 



286 MECHANICS OF ENGINEERING. 

rubbing, or sliding, force on the remainder along the under 
surface of the latter. Let the unknown intensity of this 
shearing force be X(per unit of area); then the shearing 
force on this under surface is =Xy"dx, (y",= oa in figure, 
being the horizontal width of the beam at this distance z" 
from the neutral axis of N') and takes its place with the 
other forces of the system, which are the normal stresses 

between , and portions of J and J"', the respective 

_z= z" 

total vertical shears. (The manner of distribution of J 
over the vertical section is as yet unknown ; see next arti- 
cle.) 

Putting I (horiz. compons.) = in Fig. 250, we have 

f e -p'dF— C e Z -pdF—Xy"dx=0 

J z" e J z" e 

^Xy"dx=£-£ fldF 



But from eqs. (1) and (2), p'—p = (M — if)J.=i_ dM, 
while from § 240 dM = Jdx ; 

r.Xy"dx=^ fzdFr.X =-Z- fldF .... (3) 

1 t/ z // ly" J z u v 

as the required intensity per unit of area of the shearing 
force in a surface parallel to the neutral surface and at a 
distance z" from it. It is seen to depend on the " shear " J 
and the moment of inertia / of the whole vertical section; 
upon the horizontal thickness y" of the beam at the sur- 
face in question ; and upon the integral / zdF, 

z" 
which (from § 23) is the product of the area of that part of 
the vertical section extending from the surface in question to 
the outer fibre, by the distance of the centre of gravity of that 
part from the neutral surface. 



SHEAR IX FLEXURE. 



287 



It now follows, from § 209, that the intensity (per unit 
area) of the shear on an elementary area of the vertical 
cross section of a bent beam, and this intensity we may call 
Z, is equal to that X, just found, in the horizontal section 
which is at the same distance («") from the neutral axis. 

254. Mode of Distribution of J, the Total Shear, over the Verti- 
cal Cross Section. — The intensity of this shear, Z (lbs. per 
sq. inch, for instance) has just been proved to be 



Z=X=J* I fzdF 



(4) 



iy 

To illustrate this, required the 
value of Z two inches above the neu- 
tral axis, in a cross section close to 
the abutment, in Ex. 5, § 252. Fig. 
251 shows this section. From it we 
have for the shaded portion, lying 
above the locality in question, y" — 

4 inches, and C e ~ ' zdF = (area 

J z" = 2 
of shaded portion) X (distance of 
its centre of gravity from NA) = FlG 

(12.8 sq. in.) x (3.6 in.) = 46.08 cubic inches. 

The total shear J = the abutment reaction 
while J = *. bW = I_ x 4 x (10.4) 3 = 375 biquad. inches. 
Both e/and /refer to the whole section. 




600 lbs., 



\Z= 



600x46.08 



18.42 lbs. per sq. in., 



375x4 

qui+e insignificant. In the neighborhood of the neutral 
axis, where z" = 0, we have y" = 4 and 

r e n zdF= C e zdF = 20.8 x 2.6=54.8, 
J z"=0 J 

wh__e J and I of course are the same as before. Hence 
for *" =0 



288 



MECHANICS OF ENGINEERING. 




Fig. 252 



^=^o= 21.62 lbs. per sq. in. 
At the outer fibre since J* 6 zdF-=0, z" being = e 5 Z is — 

for a beam of any shape. 

For a solid rectangular section like the 
above, Z and z" bear the same relation to 
each other as the co-ordinates of the para- 
bola in Fig. 252 (axis horizontal). 

Since in equation (4) the horizontal 
thickness, y", from side to side ef the sec- 
tion of the locality where Z is desired, 

occurs in the denominator, and since / e zdF 

J z" 

increases as z' grows numerically smaller, the following 
may be stated, as to the distribution of J, the shear, in 
any vertical section, viz.: 

The intensity (lbs. per sq. in.) of the shear is zero at 
the outer elements of the section, and for beams of ordi- 
nary shapes is greatest where the section crosses the neu- 
tral surface. For forms of cross section having thin webs 
its value may be so great as to require special investiga- 
tion for safe design. 

Denoting by Z the value of Z&t the neutral axis, (which 
=X in the neutral surface where it crosses the vertica 
section in question) and putting the thickness of the sub- 
stance of the beam = 6 at the neutral axis, we have, 



j i area above 

Z =X =-=- X< neutral axis 

I°o ( (or below) 



the dist. of its cent. 
grav.fr om that axis 



(5) 



255. Values of Zo for Special Forms of Cross Section. — From 
the last equation it is plain that for a prismatic beam the 
value of Z is proportional to J, the total shear, and hence 
to the ordinate of the shear diagram for any particular 
case of loading. The utility of such a diagram, as obtain- 



SHEAR IN FLEXURE. 



289 



ed in Figs. 234-237 inclusive, is therefore evident, for by 
locating the greatest shearing stress in the beam it 
enables us to provide proper relations between the load- 
ing and the form and material of the beam to secure safety 
against rupture by shearing. 

The table in § 210 gives safe values which the p-fr- i 
maximum Z Q in any case should not exceed. It is 
only in the case of beams with thin webs (see Figs. 
238 and 240) however, that Z is likely to need at- 
tention. 

For a Rectangle we have, Fig. 253, (see eq. 5, § 




Fig. 25a 



254) b =b, I= l / l2 bJi^ and (*zdF=tfbh . %Ji=%b¥ ' 

.\Z =X =± ji i.e., =|- (total shear) -4- (whole area) 

Hence the greatest intensity of shear in the cross-section 
is A as great per unit of area as if the total shear were 
uniformly distributed over the section. 




Fig. 254. Fig. 255. Fig. 256. 

For a Solid Circular section Fig. 254 



Fig. 257. 



J r\dF-- 

Ib Jo 


J 


2r 


Trr 2 
2 


4r_ 
ST 


_4 

= 3 ' 


J 

7TT 2 



[See§ 26Prob. 3]. 

For a Hollow Circular section (concentric circles) Fig. 
255, we have similarly, 



290 



MECHANICS OF ENGINEERING. 



&= 



J- j j~-?y kr\_jzr? 4r 2 ~] 

^(r/-r/)2(r i -r : ,)LT * 3,t T" 3* J 

__4 J(r{—r.f) 



3 n-(r/-r 2 4 )(n-*2) 

Applying this formula to Example 2 § 252, we first have 
as the max. shear J m —%P =1,735 lbs., this being the abut- 
ment reaction, and hence (putting it = (22 -5- 7)) 

7 ^^ _ 4x7xl735[64-42.8] OQA j, . 

z max. = b — = 294 lbs. per sq. m. 

3x22[256-150](4-3.5) F * 

which cast iron is abundantly able to withstand in shear- 
ing. 

For a Hollow Rectangular Beam, symmetrical about its 
neutral surface, Fig. 256 (box girder) 



— '0 



- V B (Mi s -WX&i-&*) 2 [W-WICMfc] 



The same equation holds good for Fig. 257 (I-beam with 
square corners) but then b 2 denotes the sum of the widths 
of the hollow spaces. 



256. Shearing Stress in the Web of an I-Beam. — It is usual to 
consider that, with I-beams (and box- 
beams) with the web vertical the shear J, 




in any vertical section, is borne exclusively 
by the web and is uniformly distributed 
—& — over its section. That this is nearly true 
may be proved as follows, the flange area 
being comparatively large. Fig. 258. Let 
F x be the area of one flange, and F that of 
the half web. Then since 



Fie. 258. 



K 



^^w, (!) 



SHEAR IX FLEXURE. 291 

(the last term approximate, j4 ^o being taken for the radi- 
us of gyration of F l9 ) while 

C zdF=F l ° -j-i^-L, (the first term approx.) we have 
Jo 2 4 

J C\dF 

7 __± jy,h (2F 1+ F ) whioh „_£ 

if we write (22^+^) -*- (62^+22^) =# - But b h Q is the 
area of the whole web, .-. the shear per unit area at the 
neutral axis is nearly the same as if J were uniformly dis- 
tributed over the web. E. g., with F x = 2 sq. in., and F 
= 1 sq. in. we obtain Z = 1.07 (J+h h ). 

Similarly, the shearing stress per unit area at n, the 
upper edge of the web, is also nearly equal to «/-f- b h (see 

eq., 4 (§251) for then [ F ( zdF) 1 = F x .%\ nearly, 

while I remains as before. 

The shear per unit area, then, in an ordinary I-beam is 
obtained by dividing the total shear Jhj the area of the 
web section. 

Example. — It is required to determine the proper 
thickness to be given to the web of the 15-inch wrought- 
iron rolled beam of Example 3 of §252, the height of web 
being 13 inches, with a safe shearing stress as low as 4000 
lbs. per sq. in. (the practice of the N. J. Steel and Iron 
Co., for webs), the web being vertical. 

The greatest total shear, J m9 occurring at either support 
and being equal to half the load (see table §250) we have 
with b = width of web, 

Z max.= ^-; i.e. 4000 = J??ij0 ... 6 „ = .26 inches. 
o<A> o xl3 



292 



MECHANICS OF ENGINEERING. 



(Units, inch and pound). The 15-inch light beam of the 
N. J. Co. has a web x / 2 inch thick, so as to provide for a 
shear double the value of that in the .foregoing example. 
In the middle of the span Z d = 0, since J = 0. 

257. Designing of Riveting for Built Beams. — The latter are 
generally of the I-beam and box forms, made by riveting 
together a number of continuous shapes, most of the ma- 
terial being thrown into the flange members. E. g., in fig. 
259, an I-beam is formed by riveting together, in the 
manner shown in the figure, a "vertical stem plate " or 
web, four " angle-irons," and two " flange-plates," each of 





Fig. 259. 



Fig. 260. 



these seven pieces being continuous through the whole 
length of the beam. Fig 260 shows a box-girder. If the 
riveting is well done, the combination forms a single rigid 
beam whose safe load for a given span may be found by 
foregoing rules ; in computing the moment of inertia, how- 
ever, the portion of cross section cut out by the rivet 
holes must not be included. (This will be illustrated in 
a subsequent paragraph.) The safe load having been com- 
puted from a consideration of normal stresses only, and 
the web being made thick enough to take up the max. 
total shear, J mi with safety, it still remains to design the 
riveting, through whose agency the web and flanges are 
caused to act together as a single continuous rigid mass. 
It will be on the side of safety to consider that at a given 



SHEAR Itf FLEXURE. 293 

locality in the beam the shear carried by the rivets con- 
necting the angles and flanges, per unit of length of beam, 
is the same as that carried by those connecting the angles 
and the web ("vertical stem -plate"). The amount of this 
shear may be computed from the fact that it is equal to 
that occurring in the surface (parallel to the neutral sur- 
face) in which the web joins the flange, in case the web 
and flange were of continuous substance, as in a solid I- 
beam. But this shear must be of the same amount per 
horizontal unit of length as it is per vertical linear unit in 
the web itself, where it joins the flange ; (for from § 254 Z 
=X.) But the shear in the vertical section of the web, 
being uniformly distributed, is the same per vertical linear 
unit at the junction with the flange as at any other part 
of the web section (§ 256,) and the whole shear on the ver- 
tical section of web = J, the " total shear " of that section 
of the beam. 

Hence we may state the following : 

The riveting connecting the angles with the flanges, (or 
the web with the angles) in any locality of a built beam, 
must safely sustain a shear equal to J on a horizontal length 
equal to the height of web. 

The strength of the riveting may be limited by the re- 
sistance of the rivet to being sheared (and this brings 
into account its cross section) or upon the crushing resist- 
ance of the side of the rivet hole in the plate (and this in- 
volves both the diameter of the rivet and the thickness of 
the metal in the web, flange, or angle.) In its practice the 
N. J. Steel and Iron Co. allows 7500 lbs. per sq. inch shear- 
ing stress in the rivet (wrought iron), and 12500 lbs. per 
sq. inch compressive resistance in the side of the rivet- 
hole, the axial plane section of the hole being the area of 
reference. 

In fig. 259 the rivets connecting the web with the angles 
are in double shear, which should be taken into account in 
considering their shearing strength, which is then double ; 
those connecting the angles and the flange plates are in 



294 MECHANICS OF ENGINEERING. 

single shear. In fig. 260 (box-beam) where the beam is 
built of two webs, four angles, and two flange plates, all 
the rivets are in single shear. If the web plate is very 
high compared with its thickness, vertical stiffeners in the 
form of T irons may need to be riveted upon them lat- 
erally [see § 314] . 

Example. — A built I-beam of wrought iron (see fig. 259) 
is to support a uniformly distributed load of 40 tons, its 
extremities resting on supports 20 feet apart, and the 
height and thickness of web being 20 ins. and y 2 in. re- 
spectively. How shall the rivets, which are -f in. in di- 
ameter, be spaced, between the web and the angles which 
are also y 2 in. in thickness ? Eeferring to fig. 235 we find 
that J = y 2 W = 20 tons at each support and diminishes 
regularly to zero at the middle, where no riveting will there- 
fore be required. (Units inch and pound). Near a sup- 
port the riveting must sustain for each inch of length of 
beam a shearing force of (J -r height of web) = 40000 -^ 
20 in. = 2000 lbs. Each rivet, having a sectional area of 
% n (HY = 0*60 sc l- inches, can bear a safe shear of 0.60 
X 7500 = 4500 lbs. in single shear, and .-. of 9000 lbs. in 
double shear, which is the present case. But the safe 
compressive resistance of the side of the rivet hole in 
either the web or the angle is only ^ in. x y in. X 12500 
= 5470 lbs., and thus determines the spacing of the rivets 
as follows : 

2000 lbs. -4- 5470 gives 0.36 as the number of rivets per 
inch of length of beam, i.e., they must be 1 -f- 0.36 = 2.7 
inches apart, centre to centre, near the supports; 5.4 inches 
apart at }( the span from a support; none at all in the 
middle. 

However, " the rivets should not be spaced closer than 
2}4 times their diameter, nor farther apart than 16 times 
the thickness of the plate they connect," is the rule of the 
N. J. Co. 

As for the rivets connecting the angles and flange plates* 
being in two rows and opposite (in pairs) the safe shear- 



FLEXUKE. BUILT BEAM. 



295 



ing resistance of a pair (each in single shear) is 9,000 lbs,, 
while the safe compressive resistance of the sides of the 
two rivet holes in the angle irons (the flange plate being 
much thicker) is = 10,940 lbs. Hence the former figure 
(9,000) divided into 2,000 lbs., gives 0.22 as the number of 
pairs of rivets per inch of length of the beam ; i.e., the 
rivets in one row should be spaced 4.5 inches apart, centre 
to centre, near a support ; the interval to be increased in 
inverse ratio to the distance from the middle of span, 
^bearing in mind the practical limitation just given). 

If the load is concentrated in the middle of the span, 
instead of uniformly distributed, J" is constant along each 
half-span, (see fig. 234) and the rivet spacing must accord- 
ingly be made the same at all localities of the beam. 



SPECIAL PROBLEMS IN FLEXURE. 

258. Designing Cross Sections of Built Beams. — The last par- 
agraph dealt with the riveting of the various plates ; we 
now consider the design of the plates themselves. Take 
for instance a built I-beam, fig. 261 ; one vertical stem* 




Fig. 861. 



296 MECHANICS OF ENGIKEEKING, 

plate, four angle irons, (each of sectional area = A, re- 
maining after the holes are punched, with a gravity axis 
parallel to, and at a distance = a from its base), and two 
flange plates of width = b, and thickness = t. Let the 
whole depth of girder = h, and the diameter of a rivet 
hole =f. To safely resist the tensile and compressive 
forces induced in this section by M m inch-lbs. (M m being 
the greatest moment in the beam which is prismatic) we 
have from § 239, 

<af m = — (i) 

e 
B! for wrought iron = 12,000 lbs. per sq. inch, e is = y 2 h 
while 1, the moment of inertia of the compound section, 
is obtained as follows, taking into account the fact that 
the rivet holes cut out part of the material. In dealing 
with the sections of the angles and flanges, we consider 
them concentrated at their centres of gravity (an approx- 
imation, of course,) and treat their moments of inertia 

about N as single terms in the series CdF z 2 

(see § 85). The sub tractive moments of inertia for the 
rivet holes in the web are similarly expressed ; let b = 
thickness of web. 

f 7 N for web = ^b (h—2tf—2b t' [£— t— a'] 2 
... } I N for four angles = 4A [\— t— a] 2 

{ I N for two flanges = 2(b—2f) t (^i) 2 

the sum of which makes the 7 N of the girder. Eq. (1) may 
now be written 

MJn =Is (2) 



2R' 



which is available for computing any one unknown quan- 
tity. The quantities concerned in 7 N are so numerous and 
they are combined in so complex a manner that in any 
numerical example it is best to adjust the dimensions of 
the section to each other by successive assumptions and 



FLEXURE BUILT BEAM. 297 

trials. The size of rivets need not vary much in different 
cases, nor the thickness of the web-plate, which as used 
by the N. J. Co. is " rarely less than y or more than y 
inch thick." The same Co. recommends the use of a 
single size of angle irons, viz., 3" X 3" X j£", for built 
girders of heights ranging from 12 to 36 inches, and also 
y in. rivets, and gives tables computed from eq. (2) for 
the proportionate strength of each portion of the com- 
pound section. 

Example. — (Units, inch and pound). A built I-beam 
with end supports, of span = 20 ft. = 240 inches, is to 
support a uniformly distributed load of 36 tons = 72,000 lbs. 
If y inch rivets are used, angle irons 3" X 3" X %", ver- 
tical web y 2 " in thickness, and plates 1 inch thick for 
flanges, how wide (b = ?) must these flange-plates be ? 
taking h = 22 inches = total height of girder. 

Solution. — From the table in § 250 we find that the max. 
M for this case is y Wl, where W = the total distributed 
load (including the weight of the girder) and I = span. 
.Hence the left hand member of eq. (2) reduces to 

Wl h^ 72000 x 240 x 22 

16 " B"~ ' 16 x 12000 '— 1980 

That is, the total moment of inertia of the section must 
be = 1,980 biquad. inches, of which the web and angles 
supply a known amount, since b Q = y, t =1", t'= y ,! , 
a' = 1^", A= 2.0 sq. in., a = 0.9 7 , and h = 22", are 
known, while the remainder must be furnished by the 
flanges, thus determining their width b, the unknown 
quantity. 

The effective area, A, of an angle iron is found thus : 
The full sectional area for the size given, = 3 X T A + 
2^ X y 2 = 2.75 sq. inches, from which deducting for two 
xivet holes we have 

A= 2.75—2 x y X y 2== 2.0 sq. in. 

The value a = 0.90" is found by cutting out the shape 



298 MECHANICS OF ENGINEERING. 



£3 



of two angles from sheet iron, thus : I 

and balancing it on a knife edge. (The 

gaps left by the rivet holes may be ignored, 

without great error, in findings). Hence, Fi&^ma. 

substituting we have 

I N for web =J- . ^x20 3 — 2x)4 . % [8^] 2 =282.3 
I N for four angles = 4x2x [9.10] 2 =662.5 
I N for two flanges=2(6— -f)xlx(10^) 2 =220.4(&— 1.5) 
.-. 1980= 282.3+662.5+(6— 1.5)220.4 
whence b = 4.6 + 1.5 = 6.1 inches 

the required total width of each of the 1 in. flange plates. 
This might be increased to 6.5 in. so as to equal the 
united width of the two angles and web. 

The rivet spacing can now be designed by § 257, and 
the assumed thickness of web, T / 2 in., tested for the max. 
total shear by § 256. The latter test results as follows ; 
The max. shear J m occurs near either support and = 
y 2 JF= 36,000 lbs. .*., calling b' the least allowable thickness 
of web in order to keep the shearing stress as low as 4,000 
lbs. per sq. inch, 

b' x 20" x 4000 =36000 .-. 6' =0.45" 

showing that the assumed width of ]/ 2 in. is safe. 

This girder will need vertical stiffeners near the ends, 
as explained subsequently, and is understood to be sup- 
ported laterally. Built beams of double web, or box- 
form, (see Fig. 260) do not need this lateral support. 

259. Set of Moving Loads. — When a locomotive passes over 
a number of parallel prismatic girders, each one of which 
experiences certain detached pressures corresponding to 
the different wheels, by selecting any definite position of 
the wheels on the span, we may easily compute the reac- 
tions of the supports, then form the shear diagram, and 
finally as in § 243 obtain the max. moment, M mi and the 



FLEXUKE. MOVING LOADS. 



29£ 



max. shear J mi for this particular position of the wheels. 
But the values of 3I m and J m for some other position may 
be greater than those just found. We therefore inquire 
which will be the greatest moment among the infinite 
number of (ilf m )'s (one for each possible position of the 
wheels on the span). It is evident from Fig. 236 from the 
nature of the moment diagram, that when the pressures or 
loads are detached, the M m for any position of the loads, 
which of course are in this case at fixed distances apart, 
must occur under one of the loads (i.e. under a wheel). 
We begin .*. by asking : What is the position of the set of 
moving loads when the moment under a given wheel is 
greater than will occur under that wheel in any other po- 
sition? For example, in Fig. 262, in what position of the 



t=L 



-I- <&-<&- 



0(j)!00 



p . tip. i 



ii 



-J 



Fig 262. 

loads Pi, P 2 , etc. on the span will the moment M i9 i.e., 
under P 2 , be a maximum as compared with its value under 
P 2 in any other position on the span. Let R be the resultant 
of the loads which are noiv on the span, its variable distance 
from be = x, and its Jixed distance from P 2 = a'\ while 
a, 6, c, etc., are the fixed distances between the loads 
(wheels). For any values of x , as the loading moves 
through the range of motion within which no wheel of the 
set under consideration goes off the span, and no new 



wheel comes on it, we have R l =-.~ B, 
under P 2 



and the moment 



i.e. 



=M 2 =B 1 [l-(x-a')']—PJ>-P i (b+c) 
M 2 =~(l^J^^a f x)—P,b—P i (b-{'c) 



(1) 



300 MECHANICS OF ENGINEERING. 

In (1) we have M 2 as a function of x, all the other quan- 
tities in the right hand member remaining constant as the 
loading moves ; x may vary from x =a-\-a' to 
x=l— (c-\-b— a). For a max. M 2i we put dM 2 -±dx=0, i. e. 

y(l-2x~+a')=0 .\ x (for Max M 2 ) = y 2 l+y 2 a' 

(For this, or any other value of x, cPM^dx 2 is negative, 
hence a maximum is indicated). For a max. M 2 , then, R 
must be as as far {y 2 d) on one side of the middle of the 
span as P 2 is on the other ; i.e., as the loading moves, the 
moment under a given wheel becomes a max. when that 
wheel and the centre of gravity of all the loads (then on 
the span) are equi -distant from the middle of the span. 

In this way in any particular case we may find the 
respective max. moments occurring under each of the 
wheels during the passage, and the greatest of these is the 
3I m to be used in the equation M m —R'I- : re for safe loading.* 

As to the shear J, for a given position of the wheels this 
will be the greatest at one or the other support, and 
equals the reaction at that support. When the load moves 
toward either support the shear at that end of the beam 
evidently increases so long as no wheel rolls completely 
over and beyond it. To find e/max., then, dealing with 
each support in turn, we compute the successive reactions 
at the support when the loading is successively so placed 
that consecutive wheels, in turn, are on the point of roll- 
ing off the girder at that end ; the greatest of these is the 
max. shear, J m . As the max. moment is apt to come under 
the heaviest load it may not be necessary to deal with 
more than one or two wheels in finding M m . 

Example. — Given the following wheel pressures, 

A< . . 8' . . >B< . . 5' . . >G< . . 4 . . <D 
4 tons. 6 tons. 6 tons. 5 tons, 

on one rail which is continuous over a girder of 20 ft. span, 
under a locomotive. 



* Since this may be regarded as a case of " sudden application" of a load, it is 
customary to make R' much smaller than for a dead load; from one-third to one-half 
smaller. 



FLEXURE. MOYLtfG LOADS. 



301 



1. Required the position of the resultant of A, B, and C '■> 

2. " " " " A, B, C, and I) ; 

3. " " " " B, C, and D. 

4. In what position of the wheels on the span will the 
moment under B be a max. ? Ditto for wheel C? Required 
the value of these moments and which is M m ? 

5. Required the value of J m , (max. shear), its location and 
the position of loads. 

Results.— (1.) 7.8' to right of A. (2.) 10' to right of A. 
(3.) 4.4' to right of B. (4.) Max. M B = 1,273,000 inch lbs. 
with all the wheels on ; Max. M G = 1,440,000 inch-lbs. with 
wheels B, C, and D on. (5.) J m = 13.6 tons at right sup- 
port with wheel D close to this support. 

260. Single Eccentric Load. — In the following special cases 
of prismatic beams, peculiar in the distribution of the 

loads, or mode of support, or both, 
the main objects sought are the 
-,5 values of the max. moment 3I mJ for 
(p use in the equation 

Jf m =^(see§239); 

e 

and of the max. shear J m , from 
which to design the web riveting 
in the case of an I or box-girder. 
The modes of support will be such 
that the reactions are independent 
of the form and material of the 
Fis.263. beam (the weight of beam being 

neglected). As before, the flexure is to be slight, and the 
forces are all perpendicular to the beam. 

The present problem is that in fig. 263, the beam being 
prismatic, supported at the ends, with a single eccentric 
load, P. We shall first disregard the weight of the beam 
itself. Let the span — Z x -f- ? 2 - First considering the whole 
beam free we have the reactions R, = PL + I and B 2 = 

A - 1 

Making a section at m and having Om free, x being < l 2f 
I (vert, compons.) = gives 




302 MECHANICS OF ENGINEERING. 

B 2 — J=Q, i.e., J=B 2 ; 
while from I (mom.) m =0 we have 

£- I -B,x= .\M= B 2 x=^x 
e I 

These values of J and M hold good between and G 9 J 
being constant, while M is proportional to x. Hence for 
OG the shear diagram is a rectangle and the moment dia- 
gram a triangle. By inspection the greatest M for G is 
for x = l 2 , and = P\l 2 ~- I. This is the max. M for the 
beam, since between C and B, M is proportional to the dis- 
tance of the section from B. 

r.M m =^s,nd^=I^ ... (1) 

is the equation for safe loading. 

J — B x in any section along GB, and is opposite in sign 
to what it is on OG ; i.e., practically, if a dove-tail joint 
existed anywhere on OG the portion of the beam on the 
right of such section would slide downward relatively to 
the left hand portion ; but vice versa on GB. 

Evidently the max. shear J m = B x or B 2 , as I* or \ is the 
greater segment. 

It is also evident that for a given span and given beam 
the safe load P', as computed from eq. (1) above, becomes 
very large as its point of application approaches a sup- 
port ; this would naturally be expected but not without 
limit, as the shear for sections between the load and the 
support is equal to the reaction at the near support and 
may thus soon reach a limiting value, when the safety of 
the web or the spacing of the rivets, if any, is considered. 

Secondly, considering the weight of the "beam, or any 
uniformly distributed loading, weighing w lbs. per unit of 
length of beam, in addition to P, Fig. 264, we have the 
reactions ' 

i^+f; and 2^+J" 

Let l 2 be >l x ', then for a portion Om of length a7<^, 
moments about m give 



FLEXURE. SPECIAL PROBLEMS. 



303 



x 



R 2 x -f- wx.-=Q 



pi 

— ^ , ^ 2 

i.e., on OG, M=R 2 x — y 2 wx 2 . ... . (2) 

Evidently for x = (i.e. at 0) M = 0, while for x = l 2 (i.e. 
at G) we have, putting w .= W -$- I 



M c = R 2 l 2 —*A Wlj: 



PU 2 _l M 2 



+^*25 



(3) 



J . 2 ' 2 

It remains to be seen whether a value of if may not exist 
in some section between and (7, (i.e., for a value of x 
<l 2 in eq. (2)), still greater than M c . Since (2) gives M as 
a continuous function of x between and (7, we put 
dM-^r dx = 0, and obtain, substituting the value of the con- 
stants R 2 and w, 

( max. \ pi 

R 2 — wx=0 .'. x n 1 for M or 

( min. 

This must be for M max., since d 2 M -s- dec 2 is negative 



7f'+^ 



Z. 



(4) 



mini 




Fig. 264. 



when this value of x is sub- 
stituted. If the particular 
—j value of x given by (4) is 
-P <7 2 , the corresponding value 
of if (call it i/ ) from eq. 
(2) will occur on OG and will 
be greater than M c (Dia- 
grams II. in fig. 264 show 
this case) ; but if x Q is > l 2 , 
we are not concerned with 
Rj the corresponding value of 
M, and the greatest M on OG 
would be if c . 

For the short portion BG, 
which has moment and shear 
diagrams of its own not con- 
tinuous with those for (7, it 
may easily be shown that 
M c is the greatest moment of 
any section. Hence the M 



304 MECHANICS OF ENGINEERING. 

max., or M Mi of the whole beam is either M G or M a% 
according as x n is > or < l 2 . This latter criterion may be 
expressed thus, [with l 2 — j4 I denoted by l B , the distance 
of P from the middle of the span] : 



From (eq . 4) [{^ + y,i)>i 2 ^fir [{W< (!)] 

and since from (4) and (2) 

The equation for safe loading is 

^I=M C9 when I >> h ) 
e W l x I 

and > • ... (6) 

^Lif n ,when4is < \? ) £ See ©qs. (3) and (5). 

e W \ ' for M c and M n 

If either P, W, l B , or \ is the unknown quantity sought, the 
criterion of (6) cannot be applied, and we .*. use both equa- 
tions in (6) and then discriminate between the two results. 
The greatest shear is J m ^B lt in Fig. 264, where l 2 is, 

261. Two Equal Terminal Loads, Two Symmetrical Supports 
Fig. 265. [Same case as in Fig. 231, § 238]. Neglect 
weight of beam. The reaction at each support being =P, i 
(from symmetry), we have for a free body Om with x < l x 

Pec— ^Lo .-. M=Px . . . (1) 

e 



while where x > l Y and < l x +l 



Px-P (x—li)— Sl = .-. M=Pl l . (2), 

That is, see (1), M varies directly with x between and C, 
while between G and D it is constant. Hence for safe 
loading 

—=M m Ic^I^Fk . . (3), 



FLEXTJKE. SPECIAL PROBLEMS. 



305 




Fig. 265 



The construction of the 

B moment diagram is evident 

p from equations (1) and (2). 

As for J", the shear, the 

same free bodies give, from 

I (vert. forces)=0. 

On 0(7 . J=P . . . (4) 
On CD . J=P—P=zero\$) 

(4) and (5) might also be ob- 
tained from (1) and (2) by 
writing J=d M-^rdx, but the 
former method is to be preferred in most cases, since the 
latter requires M to be expressed as a function of x while 
the former is applicable for examining separate sections 
without making use of a variable. 

If the beam is an I-beam, the fact that J is zero any- 
where on G D would indicate that we may dispense with 
a web along C D to unite the two flanges; but the lower 
flange being in compression and forming a " long column " 
would tend to buckle out of a straight line if not stayed by 
a web connection with the other, or some equivalent bracr 
ing. 

262. Uniform Load over Part of the Span. Two End Supports. 
Fig. 266. Let the load= W, extending from one support 
over a portion = c, of the span, (on the left, say,) so that 
W= wc, w being the load per unit of length. Neglect 
wei ght o f beam. For a free body Om of any length 
x < B {i.e. < c), 2 moms m =0 gives 



P T 



wx 



-B^O .\M-RiX- 



wx* 



2 " l " "-" •••(!) 

which holds good for any section on B. As for sections 
on B C it is more simple to deal with the free body m'C t 
of length 

x' < C B from which we have M=R 2 x' . . (2) 



306 



MECHANICS OP ENGINEERING. 




Fig. 266. 



which shows the moment 
curve for B C to be a straight 
line DC > tangent at D to the 



\~T parabola 0' D representing 



eq. (1.) (If there were a con- 
centrated load at B, C D 
would meet the tangent at 
D at an angle instead of co- 
inciding with it ; let the stu- 
dent show why, from the 
shear diagram). 

The shear for any value of 
x on B is : 



On OB 

while on B C 



J= i? x — ivx . , 
J= B 2 = constant 



(3) 
(4) 



The shear diagram is constructed accordingly. 



To find the position of the max. ordinate of the para- 
bola, (and this from previous statements concerning the 
tangent at the point D must occur on B, as will be seen 
and will .-. be the M m for the whole beam) we put J=0 in 
eq (3) whence 



x (for M m )= * 



W i l ~ W r- 



= C- 



w 



IV 



21 



(5) 



W 



and is less than c, as expected. [The value of B l== — Q — «) f 

L 

={wc -=-?) (I — 1), (the whole beam free) has been substi- 
tuted]. This value of x substituted in eq. (1) gives 

M^Q.—yi t) 2 . V2 . Wc .-. ^I=y 2 [i-y 2 •-* pre . .'. . (6) 

1 el 

is the equation for safe loading. 

The max. shear J m is found at and is = R^ which is 
evidently >B 2 , at C 



FLEXURE. SPECIAL PROBLEMS. 



307 



263. Uniform load Over Whole Length With Two Symmetrica 
Supports. Fig. 267. — With the notation expressed in the fig- 
ure, the following results may be obtained, after having 
divided the length of the beam into three parts for sepa- 
rate treatment as necessitated by the external forces, which 
are the distributed load W, and 
and the two reactions, each = 
y 2 W. The moment curve is 
made up of parts of three dis- 
tinct parabolas, each with its 
axis vertical. The central par- 
abola may sink below the hori- 
zontal axis of reference if the 
supports are far enough apart, 
in which case (see Fig.) the elas- 
tic curve of the beam itself becomes concave upward be- 
tween the points E and F of " contrary flexure." At each 
of these points the moment must be zero, since the radius 
of curvature is oo and M = EI -£- p (see § 231) at any sec- 
tion ; that is, at these points the moment curve crosses its 
horizontal axis. 

As to the location and amount of the max. moment M mi 
inspecting the diagram we see that it will be either at H, 
the middle, or at both of the supports B and G (which from 
symmetry have equal moments), i.e., (with I — total length,) 




Fig. 267. 



R , z either *J| %l?-in at H 

M m and.-. — \=1 T f. 

1 ™ Kl? at B and G 

21 



or 



according to which is the greater in any given case ; i.e. 
according as l 2 is > or < l Y ^/Q m 

The shear close on the left of B = wl lf while close to the 
right of B it = y 2 W — wl x . (It will be noticed that in this 
case since the beam overhangs, beyond the support, the 
shear near the support is not equal to the reaction there, 
as it was in some preceding cases.) 



308 



MECHANICS OF ENGINEERING. 



Hence J„ 



\> 



/ 2 w—w\ \ accordin g as l i <#* 



264, Hydrostatic Pressure Against a Vertical Plank. — From 
elementary hydrostatics we know that the pressure, per 
unit area, of quiescent water against the vertical side of a 
tank, varies directly with the depth, x, below the surface, 
and equals the weight of a prism of water whose altitude 
= x, and whose sectional area is unity. See Fig, 268. 



Or 


VI 


; rJrr^"" * 


N 

E 


~^=r 




^=3=^ 


R 


=«v 


/I 




■N 




Fig. 268. 



TTie plank is of rectangular cross section, its constant 
breadth, — ft, being r~ to the paper, and receives no sup- 
port except at its two extremities, and B, being level 
with the water surface. The loading, or pressure, per unit 
of length of the beam, is here variable and, by above defini- 
nition, is = w= rxb, where r = weight of a cubic unit 
(i.e. the heaviness, see § 7) of water, and x = Om — depth 
of any section m below the surface. The hydrostatic pres- 
sure on dx = wdx. These pressures for equal dx's, vary 
as the ordinates of a triangle 0B x B. 

Consider Om free. Besides the elastic forces of the ex- 
posed section m, the forces acting are the reaction B , and 
the triangle of pressure OEm. The total of the latter is 



ty* — / wk&b = rb J xdx 



rhr- 



(1) 



and the sum of the moments of these pressures about m is 
equal to that of their resultant (= their sum, since they 



x" x 



are parallel) about m, and .*. = rb --. m ~ 



FLEXUKE. SPECIAL PROBLEMS. 309 

[From (1) when x = I, we have for the total water pres- 

l 2 
sure on the beam W x = yb — and since one-third of this 

will be borne at we have R =*}£ ybl 2 .] 

Now putting I{ moms, about the neutral axis of ra)=0, 
for Om free, we have 



R x—JV % . ~—^I=0 .-. M= %yb (l 2 x—tf) 



(2) 



(which holds good from x = to x = Z). From I (horiz. 
forces) = we have also the shear 

J=B —W x =y 6 yb (V—3x>) (3) 

as might also have been obtained by differentiating (2), 
since J — dM -r- dx. By putting J = (§ 240, corollary) 
we have for a max. ill, x = I 4- Vg, which is less than I 
and hence is applicable to the problem. Substitute this 
in eq. 2, and reduce, and we have 

R f I ^ . R'l 1 1 

~=M m , i.e. — =g ■ 7|- • r» (4) 

as the equation for safe loading. 

265. Example. — If the thickness of the plank is h, re- 
quired h = ?, if i2' is taken = 1,000 lbs. per sq. in. for 
timber (§ 251), and I = 6 feet. For the inch-pound-second 
system of units, we must substitute R' = 1,000 ; I = 72 
inches ; y = 0.036 lbs. per cubic inch (heaviness of water 
in this system of units); while I =bh s -=- 12, (§ 247), and e 
— y 2 h. Hence from (4) we have 

1000 W 0.0366 x 72 3 M ' fl . Qr7 . 
^ — -tt= n 7 - , .*. #=5.16 ,\ A = 2.27 m. 

It will be noticed that since a? for M m = Z -4- V 3, and not 
^3 Z, ilf m does not occur in the section opposite the resul- 
tant of the water pressure ; see Fig. 268. The shear curve 
is a parabola here ; eq. (3). 



wdx f j,+ d<i 



lii 



N N 



310 MECHANICS OF ENGINEERING. 

266. The Four x-Derivatives of the Ordinate of the Elastic Curve 
— If y = func. (x) is the equation of the elastic curve for 
any portion of a loaded beam, on which portion the load 
per unit of length of the beam is w = either zero, (Fig. 
234) or = constant, (Fig. 235), or = a continuous func. (x) 
(as in the last §), we may prove, as fol- 
lows, that w = the ^-derivative of the 
shear. Fig. 269. Let N and N' be two 
consecutive cross-sections of a loaded 
beam, and let the block between them, 
bearing its portion, wdx, of a distributed 
\ *-- dz -A " " load, be considered free. The elastic 
forces consist of the two stress-couples 
Fig. 269. (tensions and compressions) and the two 

shears, J" and J + dJ, dJ being the shear-increment conse- 
quent upon x receiving its increment dx. By putting 
^(vert. components) -Owe have 

J-\-dJ- — wdx — J=0 .\ w==— — 

dx 

Q. K D. But J itself = dM + dx, (§ 240) and 

M = [d 2 y -T- dx 2 ] EL By substitution, then, we have the 

following relations : 

y =func.(x) = ordinate at any point of the elastic curve (1) 
-Jr-~ « = slope at any point of the elastic curve •. . (2) 

d 2 v 
EI -t4" = M = ordinate (to scale) of the moment curve (3) 
dx 

,. r&y ., , T { the ordinate (to scale) ) //n 

EI % = * ' I ° f the shear W am i • • {) 

( the load per unit of length | 
EI —^ = w = \ of beam = ordinate (to scale) V . (5) 
drf ( of a curve of loading. ) 

If, then, the equation of the elastic curve (the neutral line 
of the beam itself ; a reality, and not artificial like the 



FLEXURE. SPECIAL PROBLEMS. 311 

other curves spoken of) is given ; we may by successive 
differentiation, for a prismatic and homogeneous beam so 
that both E and / are constant, find the other four quan- 
tities mentioned. 

As to the converse process, (i.e. having given w as a 
function of x, to find expressions for J 9 M and y as func- 
tions of x) this is more difficult, since in taking the 
ce-anti-derivative, an unknown constant must be added and 
determined. The problem just treated in § 264, however, 
offers a very simple case since w is the same function of 
x, along the tuhole beam, and there is therefore but one elas- 
tic curve to be determined. 

We .*. begin, numbering backward, with 

tpt &11 7 ( since w = rbx. see ] /r * 

^ 7 d! = ~ rfa ! last § and Fig. 268 f • ■ • ( fo ) 

[N. B. — This derivative (dJ~-dx) is negative since dJ and 
dx have contrary signs.] 

.-. (shear=)E I ( ^y=—rb ~+Const 
ax 6 2 

But writing out this equation for x=0, i.e. for the point 
0, where the shear=i? , we have B<f= + Const .'. Const = 
R 0> and hence write 

^l^^l^ " ( Shear) * (4a) 
Again taking the x- anti-derivative of both sides 

(Moment=)IJI c pt=— r b^-\-B x+(Const=0) . (3a) 
ax? o 

[At 0, ^=0 also M, .*. Const. =0]. Again, 

At 0, where x=0 dy -^dx^a i) =the unknown slope of the 
elastic line at 0, and hence C f =FIa 



312 MECHANICS OF ENGINEERING. 

passing now to y itself, and remembering that at 0, both 
y .and x are zero, so that the constant, if added, would= 
zero, we obtain (inserting the value of B from last §) 

My=- r b^+ r bl*^ + EIa x . (1«) 

the equation of the elastic curve. This, however, contains 
the unknown constant « =the slope at 0. To determine 
a Q .write out eq. (la) for the point B, Fig. 268, where x is 
known to be equal to I, and y to be = zero, solve for « , 
and insert its value both in (la) and (2a). To find the 
point of max. y (i.e., of greatest deflection) in the elastic 
curve, write the slope, i.e. dy -=- dx, = zero [see eq. 2a] and 
solve for x ; four values will be obtained, of which the one 
lying between and I is obviously the one to be taken. 
This value of x substituted in (la) will give the maximum 
deflection. The location of this maximum deflection is 



neither at the centre of action of the load 



HO 



nor at the section of max. moment (x =Z-v-^/ 3.) 

The qualities of the left hand members of equations (1) 
to (5) should be carefully noted. E. g., in the inch-pound- 
second system of units we should have : 

1. y (& linear quantity) = (so many) inches. 

2. dy-7-dx (an abstract number) = (so many) abstract 
unifs. 

(sSl M (a moment) = (so many) inch-pounds. 

4. J (a shear, i.e., force) = (so many) pounds. 

5. iv (force per linear unit) = (so many) pounds per run- 
ning inch of beam's length. 

As to the quantities E, and /, individually, E is pounds 
per'sq. in., and /has four linear dimensions, i.e. (so many) 
bi-quadratic inches. 

v :) 



FLEXURE SPECIAL PROBLEMS. 313 

267. Resilience of Beam With End Supports. — Fig. 270. If a 
9g mass whose weight is G (G large com- 

. \ h , pared with that of beam) be allowed to 

. ^. J 3 b > ^ a ^ f ree ly through a height = h upon 
3 7 ( S the centre of a beam supported at its 

fcj ' p « extremities, the pressure P felt by the 

fig. 270. beam increases from zero at the first 

instant of contact up to a maximum P ra , as already stated 

in §233a, in which the equation was derived, d m being 

small compared with h, 

1 P 2 7 3 

The elastic limit is supposed not passed. In order that 
the maximum normal stress in any outer fibre shall at most 
be=i? / , a safe value, (see table §251) we must put 

/?' T PI 

=—j- [according to eq. (2) §241,] i.e. in equation (a) 

above, substitute P m = [4 R'T\ -r-le, which gives 

E»2 77 r>/2 7,2 r>/2 7,2 

having put I=Fk 2 (h being the radius of gyration §85) 
and Fl= V the volume of the (prismatic) beam. From 
equation (b) we have the energy, Gh, (in ft. -lbs., or inch- 
lbs.) of the vertical blow at the middle which the beam of 
Fig. 270 will safely bear, and any one unknown quantity 
can be computed from it, (but the mass of G should not 
be small compared with that of the beam.) 

The energy of this safe impact, for two beams of the 
same material and similar cross-sections (similarly placed), 
is seen to be proportional to their volumes; while if further- 
more their cross-sections are the same and similarly 
placed, the safe Gh is proportional to their lengths. (These 
same relations hold good, approximately, beyond the elas- 
tic limit.) 

It will be noticed that the last statement is just the con- 



314 



MECHANICS OF ENGINEERING. 



verse of what was found in §245 for static loads, (the 
pressure at the centre of the beam being then equal to 
the weight of the safe load) ; for there the longer the beam 
(and .*. the span) the less the safe load, in inverse ratio. 
As appropriate in this connection, a quotation will be 
given from p. 186 of " The Strength of Materials and 
Structures," by Sir John Anderson, London, 1884, viz.: 

" It appears from the published experiments and state- 
ments of the Railway Commissioners, that a beam 12 feet 
long will only support y 2 of the load that a beam 6 feet 
long of the same breadth and depth will support, but that 
it will bear double the weight suddenly applied, as in the 
case of a weight falling upon it," (from the same height, 
should be added) ; " or if the same weights are used, the 
longer beam will not break by the weight falling upon it 
unless it falls through twice the distance required to frac- 
ture the shorter beam." 

268. Combined Flexure and Torsion. Crank Shafts. Fig. 271. 
Let OiB be the crank, and N0 X the portion projecting 

beyond the nearest bearing 
N. P is the pressure of the 
connecting-rod against the 
crank -pin at a definite in- 
stant, the rotary motion be- 
ing uniform. Let a== the 
perpendicular dropped from 
the axis 00 x of the shaft 
upon P, and 1= the distance 
of P, along the axis X from 
the cross-section NmN' of the 
Let NN' be a diameter of this 
In considering the portion 
NOiB free, and thus exposing the circular section NmN\ 
we may assume that the stresses to be put in on the ele- 
ments of this surface are the tensions (above NN 9 ) and 
the compressions (below NN') and shears ~| to NN' 9 due 
to the bending action of P ; and the shearing stress tan. 




shaft, close to the bearing 
section, and parallel to 



a. 



FLEXURE. SPECIAL PROBLEMS. 



315 



gent to the circles which have as a common centre, and 
pass through the respective clF's or elementary areas, 
these latter stresses being due to the twisting action of P. 
In the former set of elastic forces let p = the tensile 
stress per unit of area in the small parallelopipedical ele- 
ment m of the helix which is furthest from NN' (the neu- 
tral axis) and 1= the moment of inertia of the circle about 
NN'; then taking moments about NN' for the free body, 
(disregarding the motion) we have as in cases of flexure 
(see §239) 

pi ™ _ . _ Plr 



PI; 



p- 



(«) 



[None of the shears has a moment about NN'.] Next 
taking moments about 00 Xi (the flexure elastic forces, both 
normal and shearing, having no moments about 00i) we 
have as in torsion (§216) 



*^=Pa; i.e.,p s =-—- 



(P) 



in which p s is the shearing stress per unit of area, in the 
torsional elastic forces, on any outermost dF, as at m; 
and i~ p the polar moment of inertia of the circle about its 
centre 0. 

Next consider free,, in Fig. 272, a small parallelopiped 
taken from the helix at m (of Fig. 271.) The stresses [see 
§209] acting on the four faces |— to the paper in Fig. 272 
are there represented, the dimensions (infinitesimal) being 
n || to NN, b || to 0Oi, and d H to the paper in Fig. 272. 



/pud 


T 


^P s nd 


p s nd' N - 

4 T 


P/*J 


i 


K. e 

- 11 "(X / 


P s oa 


^SSsfii b 1 






pnd 


pnd 


5 53 


i H 

f b M 

~p s tid 




/ h 

qcd~* 



Fig. 272. 



Fig. 273. 



316 MECHANICS OF ENGINEERING. 

By altering the ratio of b to n we may make the angle 6 
what we please. It is now proposed to consider free the 
triangular prism, GHT 3 to find the intensity of normal 
stress g, per unit of area, on the diagonal plane GH, (of 
length = c,) which is a bounding face of that triangular 
prism. See Fig, 273. By writing I (compons. in direc- 
tion of normal to GH)=0, we shall have, transposing, 

qcd—pnd sin 6+p s bd sin 6-\-p s nd cos ; and solving for q 
q=p -. sin 6+p s r-sin0+-. cos 6j ; . (1) 

but n : c= sin and b : c= cos 6 .*. 

q=p sin 2 d-\-pJ2 sin cos 6 . . (2) 

This may be written (see eqs. 63 and 60, O. W. J. Trigo- 
nometry) 

g=i^9(l_cos2#)+l> s sin20 . . (3) 

As the diagonal plane GH is taken in different positions 
(i.e., as varies), this tensile stress q (lbs. per sq. in. for 
instance) also varies, being a function of 6, and its max. 
value may be >J9. To find 6 for q max. we put 

g| , =p sin 2d+2p s cos 26, . (4) 

2p s 
= 0, and obtain : tan [2(0 for g max)]= — — - . . • (5) 

Call this value of 6, 6'. Since tan 2d' is negative, 26' lies 
either in the second or fourth quadrant, and hence 

8in2 ^ ± ^¥i andcos2 ^^fe (6) 

[See equations 28 and 29 Trigonometry, O. W. J.] The 



FLEXURE. CRANK SHAFT. 3j" 

upper signs refer to the second quadrant, the lower to the 
fourth. If we now differentiate (4), obtaining 

gj[= 2p cos 20-4p B sin 20 . (7) 

we note that if the sine and cosine of the [20'] of the 2nd 
quadrant [upper signs in (6)] are substituted in (7) the re- 
sult is negative, indicating a maximum ; that is, q is a max- 
imum for 0= the 0' of eq. (6) ichen the upper signs are taken 
(2nd quadrant). To find q max., then, put 0' for in (3) 
substituting from (6) (upper signs). We thus find 

q max =^[p+Vj? 2 +4p s 2 .] . . (8) 

A similar process, taking components parallel to GH> 
Fig. 273, will yield q s max., i.e., the max. shear per unit of 
area, "vhich for a given p and p s exists on the diagonal 
plane GH in any of its possible positions, as varies. 
This max. shearing stress is 

g s max =j£yy+4p B 2 . . (9) 

In the element diametrically opposite to m in Fig. 271, p 
is compression instead of tension ; q maximum will also 
be compression but is numerically the same as the q max. 
of eq. 8. 

269. Example.— In Fig. 271 suppose P=2 tons = 4,000 
lbs., a=6 in., 1=5 in., and that the shaft is of wrought 
iron. Eequired its radius that the max. tension or com- 
pression may not exceed R'= 12,000 lbs. per sq. in.; nor the 
max. shear exceed S' = 7,000 lbs. per sq. in. That is, we 
put <7= 12,000 in eq. (8) and solve for r : also ^ s = 7,000 in 
(9) and solve for r. The greater value of r should be 
taken. From equations (a) and (b) we have (see §§ 219 and 
247 for I p and 1) 



318 



MECHANICS OF ENGIKEERIKGk 



4PZ , 2Pa 
p= — 5- and Ps- 



tzt 3 



tit* 



which in (8) and (9) give 



max. q=V2~ [4:l+V{4:iy+4:(2ay] 



(8a) 



and 



max. q, 



7ZT 



; V(4Z) 2 +4(2a) 2 



(9a) 



With max. g=12,000, and the values of P, a, and I, already 
given, (units, inch and pound) we have from (8a), ^=2.72 
cubic inches .*. r=1.39 inches. 

Next, with max. ^ s = 7,000 ; P, a, and I as before ; from 
(9a), r 3 =2.84 cubic inches .\ r=lAl inches. 

The latter value of r, 1.41 inches, should be adopted. It 
is here supposed that the crank-pin is in such a position 
(when P= 4,000 lbs., and a=6 in.) that q max. (and q s 
max.) are greater than for any other position ; a number 
of trials may be necessary to decide this, since P and a are 
different with each new position of the connecting rod. If 
the shaft and its connections are exposed to shocks, JB! and 
S' should be taken much smaller. 

270. Another Example of combined torsion and flexure is 
,. shown in Fig. 274. The 

p b work of the working force 
P 2 ( vertical cog-pressure) is 
B expended in overcoming the 
resistance (another vertical 
cog-pressure) Q x . 

That is, the rigid body 
consisting of the two wheels and shaft is employed to 
transmit power, at a uniform angular velocity, and since 
it is symmetrical about its axis of rotation the forces act- 
ing on it, considered free, form a balanced system. (See 
§ 114). Hence given P { and the various geometrical quan- 




Fig. 274. 



FLEXUKE. CRANK SHAFT. 319 

tities a l9 b it etc., we may obtain Q u and the reactions P and 
P B , in terms of P±. The greatest moment of flexure in the 
shaft will be either P l ly at C; or P B l 3f at D. The portion 
CD is under torsion, of a moment of torsion =P 1 a 1 = Q x b lm 
Hence we proceed as in the example of § 269, simply put- 
ting PJj X (or P B l B , whichever is the greater) in place of PI, 
and P\0\ in place of Pa. "We have here neglected the 
weight of the shaft and wheels. If Qi were an upward ver- 
tical force and hence on the same side of the shsit as P lf 
the reactions P and P B would be less than before, and one 
or both of them might be reversed in direction. 



320 MECHANICS OF ENGINEERING. 



CHAPTEE IV. 



FLEXURE, CONTINUED, 



CONTINUOUS GIRDERS. 



271. Definition. — A continuous girder, for present pur- 
poses, may be defined to be a loaded straight beam sup- 
ported in more than two points, in which case we can no 
longer, as heretofore, determine the reactions at the sup- 
ports from simple Statics alone, but must have recourse 
■to the equations of the several elastic curves formed by its 
neutral line, which equations involve directly or indirect- 
ly the reactions sought ; the latter may then be found as 
if they were constants of integration. Practically this 
amounts to saying that the reactions depend on the man- 
ner in which the beam bends ; whereas in previous cases, 
with only two supports, the reactions were independent of 
the forms of the elastic curves (the flexure being slight, 
however). 

As an Illustration, if the straight beam of Fig. 275 is placed 
on three supports 0, B, and C, at the same level, the 
reactions of these supports seem at first sight indeterm- 
inate ; for on considering the $ .\ h 

whole beam free, we have three CI3I^II^L____d_^__6 

unknown quantities and only B /J^- — p J\o S — ^2\ 

two equations, viz : 2 (vert. F ig. 275. 

compons.) = and 2 (moms, about some points = 0. If 
now be gradually lowered, it receives less and less pres- 



FLEXURE. CONTINUOUS GIRDERS. 321 

sure, until it finally reaches a position where the beam 
barely touches it ; and then O's reaction is zero, and B and 
C support the beam as if were not there. As to how 
low must sink to obtain this position, depends on the 
stiffness and load of the beam. Again, if be raised 
above the level of B and G it receives greater and greater 
pressure, until the beam fails to touch one of the other 
supports. Still another consideration is that if the beam 
were tapering in form, being stiffest at 0, and pointed at 
B and (7, the three reactions would be different from their 
values for a prismatic beam. It is therefore evident that 
for more than two supports the values of the reactions de- 
pend on the relative heights of the supports and upon the 
form and elasticity of the beam, as well as upon the load. 
The circumstance that the beam is made continuous over 
the support 0, instead of being cut apart at into two 
independent beams, each covering its own span and hav- 
ing its own two supports, shows the significance of the 
term " continuous girder." 

All the cases here considered will be comparatively 
simple, from the symmetry of their conditions. The 
beams will all be prismatic, and all external forces (i.e. 
loads and reactions) perpendicular to the beam and in the 
same plane. All supports at the same level. 

272. Two Equal Spans; Two Concentrated Loads, One in the Mid- 
dle of Each Span. Prismatic Beam. — Fig. 275. Let each half - 
span = ]/ 2 I. Neglect the weight of the beam. Required 
the reactions of the three supports. Call them P Bf P and 
P c . From symmetry P B = P c , and the tangent to the 
elastic curve at is horizontal ; and since the supports 
are on a level the deflection of (and B) below O's tangent 
is zero. The separate elastic curves OB and DC have a 
common slope and a common ordinate at D. 

For the equation of OD, make a section n anywhere be- 
tween and D, considering nG a free body. Fig. 276 (a) 



322 



MECHANICS OF ENGINEEKING. 




-* 1 



X 



(Si 



with origin and axis as there indicated. From I (moms 
about neutral axis of n) = we have (see § 231) 

EI^=P(%l-x)-P<{l^x) . . . 



ElijL =J>(^l x ^j-PJlx-^+(O=0) 



(1 
(2) 



The constant — 0, for at both x, and dy -f- dx, = 0. 
Taking the x-anti-derivative of (2) we have 

EIy=P^J^-Pj^J^ . . (3) 

Here again the constant is zero since at 0,x and y both =0. 
(3) is the equation of OD, and allows no value of x <0 
or>-. It contains the unknown force P c . 

For the equation of DC, let the variable section n be made 
anywhere between D and (7, and we have (Fig. 276 (b) ; x 
may now range between y 2 l and I) 



#&-**-* 



EI dy = _ P / ia! _A + C / 

da; C V 2 / 



(4) 

(5)' 



To determine C, put cc = y 2 l both in (5)' and (2), and 
equate the results (for the two curves have a common 
tangent line at D) whence C = }i PI 2 

dy T/ -nil r»/7„ x*\ 



^2=* '"•Mr) 



(5) 



FLEXUHE. CONTINUOUS GIEDEES. 323 

Hence Ely = tf PPx—P c f^— Jj+(7" . . (6)' 

At D the curves have the same y, hence put x = 1 in the 
right hand member both of (3) and (6)', equating results, 
and we derive G"= — A PI 3 

My=ysPP^P^^-±PF . . (6) 

which is the equation of DC, but contains the unknown 
reaction P c . To determine P c we employ the fact that O's 
tangent passes through C, (supports on same level) and 
hence when x = I in (6), y is known to be zero. Making 
these substitutions in (6) we have 

From symmetry P B also = -AP, while P must = ~P, 
since P B + P + P^ = 2 P (whole beam free). [Note. — 
If the supports were not on a level, but if, (for instance) 
the middle support were a small distance = h below 
the level line joining the others, we should put x = I and 
y = — h in eq. (6), and thus obtain P B = P c — ~^ & P + 

3EI-£, which depends on the material and form of the 

6 

prismatic beam and upon the length of one span, (whereas 
with supports all on a level, P B — P c = A P is independent 
of the material and form of the beam so long as it is ho- 
mogeneous and prismatic.) If P , which would then = 
?| P — 6 EI (Jiq-^-W), is found to be negative, it shows that 
requires a support from above, instead of below, to 
cause it to occupy a position h below the other supports, 
i.e. the beam must be " latched down " at 0.] 

The moment diagram of this case can now be easily con- 
structed ; Fig. 277. For any free body nC 9 n lying in DG t 
we have - , 



324 



MECHANICS OF ENGINEERING. 



I 









M^UPx, 



1—4 



i.e., varies directly as x, un- 

2 j"" I £ £P ^ x P asses D when, for any 
point on DO, 



L 



a-J 



;<uu^^ 



Oil 



i!f=V 16 P a; -P( a; -i) 



which is =0, (point of in- 
flection of elastic curve) 
1 for x= 8 / n I (note that x is 
measured from G in this 



-k pl 



Fig. 277. 

figure) and at 0, where x= I, becomes 

••• K—f 2 Pl; M G =0; M B =^Ph, andif c =0 
Hence, since M max. = ^Pl t the equation for safe loading 
is 

t?'T a 

. . . . (7) 



M=^pi 



e 32 

The shear at G and anywhere on GD=^P, while on. DO 
it =~P in the opposite direction 

.-.j- m =;}P .... (8) 

The moment and shear diagrams are easily constructed, 
as shown in Fig. 277, the former being svmmetrical about 
a vertical line through 0, the latter about the point 0" 
Both are bounded by right lines. 

273. Two Equal Spans. Uniformly Distributed Load Over 

Whole Length. Prismatic Beam, 
c —Fig. 278. Supports B, 0, 
B i 1 J JJJ-LLLU I l| I C, on a level. Total load 

(^ pi — ~j^T ^j = 2 W= 2wl and may include 
|p I M( £r z) I that of the beam ; w is con- 
I 1 U \ \ II stant. As before, from sym- 
metry P B =P C , the unknown 
p c | reactions at the extremi- 
st. 278. ties. 



W = wZ 



W=«>Z 




FLEXURE. CONTINUOUS GIRDERS. 



325 



Let On=x ; then with n G free, I moms, about n= gives 

FI^ = tv(l-x)( l ^)-P c (l-x)= ™[P-2lx+x*]-P c (l-x) (1) 

... eA=% [fc-ZB»+£]-P c [&- ^] + [Const=0] (2) 
ax z o 2 

[Const. =0 for at both dy-^dx the slope, and #, are =0] 

•■• %%= |[^^ 2 -K^+Vi2^]-A[^^ 2 -^]+(C=0) (3) 

[Const. =0 for at both x and?/ are =0]. Equations (1), 
(2), and (3) admit of any value of x from to I, i.e., hold 
good for any point of the elastic curve OG, the loading on 
which follows a continuous law (viz. : w— constant). But 
when x=l, i.e., at (7, y is known to be equal to zero, since 
0, B and G are on the axis of X, (tangent at 0). With 
these values of x and y in eq. (3) we have 

0= JL . t-%PJP .'. P=3/ 8 wl=3/sW 

... P B =3/ S W and P =2 JT— 2P =!° W 



The Moment and Shear Diagrams can now be formed since 

all the external forces are 
±w i known. In Fig. 279 meas- 
ure x from G. Then in any 
section n the moment of the 
"stress-couple " is 

wx 2 




M=VrWx- 



(1) 



which holds good for any 
value of x on CO, i.e., from 
x=0 up to x—l. By inspec- 
Fig. 279. tion it is seen that for x=0, 

M=0 ; and also for x=ffl, M=0, at the inflection point G, 
beyond which, toward 0, the upper fibres are in tension 



326 MECHANICS OF ENGINEERING. 

the lower in compression, whereas between G and G they 
are vice versa. As to the greatest moment to be found on 
GG, put dM^-dx=0 and solve for x. This gives 

y& W—wx=0 ,\ \x for M max.] = y& . (2) 

which in eq. (1) gives 

M N (a,tN 9 see figure)=+4 JFZ . . (2) 

But this is numerically less than M (= — y& Wl) hence the 
stress in the outer fibre at being 

T/ Wle /QX 

Po=H—j-> ( 3 ) 

the equation for safe loading is 

*lL=#WI .... (4) 

the same as if the beam were cut through at 0, each half, 
of length I, retaining the same load as before [see § 242 eq. 
(2)]. Hence making the girder continuous over the mid- 
dle support does not make it any stronger under a uni- 
formly distributed load ; but it does make it considerably 
stiffer. 

As for the shear, J, we obtain it for any section by tak- 
ing the ^-derivative of M in eq. (1), or by putting ^(ver- 
tical forces) =0 for the free body nG, and thus have for 
any section on GO 

J=y£W—wx ... (5) 

J" is zero for x = y&l (where M reaches its calculus maxi- 
mum M N ; see above) and for x=l it = — $/% W which is nu 
merically greater than ^ W, its value at 6". Hence 

J m = s/ 8 W .... (6) 



FLEXURE. CONTINUOUS GIRDERS. 



327 



The moment curve is a parabola (a separate one for each 
span), the shear curve a straight line, inclined to the hor- 
izontal, for each span. 

Problem. — How would the reactions in Fig. 278 be 
changed if the support were lowered a (small) distance 
h below the level of the other two ? 



274. Prismatic Beam Fixed Horizontally at Both Ends (at 
Same Level). Single Load at Middle. — Fig. 280. [As usual 

the beam is understood to 
be homogeneous so that E 
is the same at all sections]. 
The building in, or fixing, 
of the two ends is supposed 
to be of such a nature as to 
cause no horizontal con- 
straint ; i.e., the beam does 




V -*- 1 - 



B 

Fig. 280. 

not act as a cord or chain, in any manner, and hence the 
sum of the horizontal components of the stresses in any 
section is zero, as in all preceding cases of flexure. In 
other words the neutral axis still contains the centre of 
gravity of the section and the tensions and compressions 
are equivalent to a couple (the stress-couple) whose mo- 
ment is the " moment of flexure." 

If the beam is conceived cut through close to both wall 
faces, and this portion of length=Z, considered free, the 
forces holding it in equilibrium consist of the downward 
force P (the load) ; two upward shears J and J c (one at 
each section) ; and two " stress-couples " one in each sec- 
tion, whose moments are M and M c . From symmetry we 
know that J =J C , and that M =3f c . From I 7=0 for the 
free body just mentioned, (but not shown in the figure), 
and from symmetry, we have J = }4 P and J c = x / 2 P ; but 
to determine M n and M c , the form of the elastic curves 
B and B G must be taken into account as follows : 

Equation of OB, Fig. 280. I [mom. about neutral axis 
of any section n on B] = (for the free body nC which 



328 



MECHANICS OF ENGINEERING. 



lias a section exposed at each end, n being the variable 
section) will give 



EI 



d 2 y 



dx 2 



p(y 2 i-x) + M—y 2 P(i—x) 



(i) 



[Note. In forming this moment equation, notice that 
M c is the sum of the moments of the tensions and com- 
pressions at about the neutral axis at n, just as much as 
about the neutral axis of 0; for those tensions and com- 
pressions are equivalent to a couple, and hence the sum of 
their moments is the same taken about any axis whatever 
"1 to the plane of the couple (§32).] 

Taking the x-anti-derivative of each member of (1), 



dx 



■F(y 2 1 x—y 2 3*)+ m c x—y 2 p(i x —y 2 x 2 ) 



(2) 



(The constant is not expressed, as it is zero). Now from 
symmetry we know that the tangent-line to the curve B 
at B is horizontal, i.e., for x=y 2 l, dy-i-dx=0, and these 
values in eq. (2) give us 

0= y Pl 2 + y 2 M c l—^Pl 2 ; whence M C =M = % PI . (3) 

Safe Loading. Fig. 281. Having now all the forces which 
act as external forces in straining the beam 00, we are 
ready to draw the moment diagram and find M m . For con- 
venience measure x from C. For the free body nC, we 
have [see eq. (3)] 

y 2 px-m +2*=o .-. 3i=y s pi-y 2 Px ... (4) 

Eq. (4) holds good for any 
2 J section on OB. By put- 
fT** ting x=0 we have M=3I C = 
PI; layof£HG'=M to 
scale (so many inch-pounds 
moment to the inch of pa- 
per). At B, for x~y I, 
3I B = — y PI ; hence lay 
off B'D=y PI on the op- 
no. 28i. posite side of the axis 0'G f 




FLEXURE. CONTINUOUS GIRDERS. 



329 



from H C, and join DH. DK, symmetrical with DH about 
B'D, completes the moment curves, viz.: two right lines. 
The max. if is evidently =^$ PI and the equation of safe 
loading 



VI^APl 



(5) 



Hence the beam is twice as strong as if simply supported 
at the ends, under this load ; it may also be proved to be 
four times as stiff. 

The points of inflection of the elastic curve are in the 
middles of the half-spans, while the max. shear is 



J m = MP 



(6) 



275. Prismatic Beam Fixed Horizontally at Both Ends [at Same 
Level], Uniformly Distributed Load Over the Whole Length. 

Fig. 282. As in the preceding problem, we know from 
symmetry that J = J c = y 2 W= yi wl, and that M =M C , and 
determine the latter quantities by the equation of the 
curve 0(7, there being but one curve in the present in- 
stance, instead of two, as there is no change in the law of 
loading between and C. With nO free, 2 (mom u )=0 
gives 



dx 2 



y 2 Wx + m + 



vxx? 



and .-. EA=—y 2 W^ + M o x+^+[C=0] 
ax 2 o 




(i) 



(2) 



Fig. 282. 



330 



MECHANICS OF ENGINEERING. 



The tangent line at being horizontal we have for x=0,-j — 

0, .*. (7=0. But since the tangent line at C is also hori- 
zontal, we may for x=l put dy-^-dx=0 t and obtain 



0=—}£WP+m+}4u&; whence M =±Wl 



(3) 



as the moment of the stress-couple close to the wall at 
and at C. 

Hence, Fig. 283, the equation of the moment curve (a 
single continuous curve in this case) is found by putting 
I (mom n )=0 for the free body nO, of length x, thus, 
obtaining 



i.e., 




Fig. 283. 



e 



+ y 2 Wx-M 



wx 2 



ivx 2 



m=±wi+ ^--y 2 Wx 



(4) 



an equation of the second degree, indicating a conic. At 0, 
M=M Q of course,= A- Wl ; at B by putting x=y 2 I in (4), we 
have M B = — 2 \ Wl, which is less than M , although 3I B is the 
calculus max. (negative) for 31, as may be shown by writ- 
ing the expression for the shear (J=j4 W—wx) equal to- 
zero, etc. 



FLEXUEE. COXTIXTTOUS GIKDEES. 331 

Hence M m =~Wl, and the equation for safe loading is 

^=4 m (5) 

Since (with this form of loading) if the beam were not 
built in but simply rested on two end supports, the equa- 
tion for safe loading would be [B'I-r-e] = *& Wl, (see §242), 
it is evident that with the present mode of support it is 50 
per cent, stronger as compared with the other ; i.e., as re- 
gards normal stresses in the outer elements. As regards 
shearing stresses in the web if it has one, it is no stronger* 
since J m = j4 Win both cases. 

As to stiffness under the uniform load, the max. deflec- 
tion in the present case may be shown to be only ~ of that 
in the case of the simple end supports. EH£ 

It is noteworthy that the shear diagram in Fig. 283 is 
identical with that for simple end supports §242, under 
uniform load ; while the moment diagrams differ as fol- 
lows : The parabola KB' A, Fig. 283, is identical with that 
in Fig. 235, but the horizontal axis from which the ordi- 
nates of the former are measured, instead of joining the 
extremities of the curve, cuts it in such a way as to have 
equal areas between it and the curve, on opposite sides 

i.e., areas [KC'IP+AG'0']=xrQS> H'G'B' 

In other words, the effect of fixing the ends horizontally^ 
is to shift the moment parabola upward a distance = M c 
(to scale), = i Wl, with regard to the axis of reference, 
O'B', in Fig. 235. 

276. Remarks. — The foregoing very simple cases of con- 
tinuous girders illustrate the means employed for deter- 
mining the reactions of supports and eventually the max. 
moment and the equations for safe loading and for deflec- 
tions. When there are more than three supports, with 
spans of unequal length, and loading of any description, 
the analysis leading to the above results is much more 
complicated and tedious, but is considerably simplified 



332 MECHANICS OF ENGINEERING. 

and systematized by the use of the remarkable theorem of 
three moments, the discovery of Clapeyron, in 1857. By 
this theorem, given the spans, the loading, and the vertical 
heights of the supports, we are enabled to write out a rela- 
tion between the moments of each three consecutive sup- 
ports, and thus obtain a sufficient number of equations to 
determine the moments at all the supports [p. 641 Kankine's 
Applied Mechanics.] From these moments the shears 
close to each side of each support are found, then the 
reactions, and from these and the given loads the moment 
at any section can be determined ; and hence finally the 
max. moment 3I m , and the max. shear J m . 

The treatment of the general case of continuous girders 
by graphic methods, however, is comparatively simple, 
and its presentation is therefore deferred, § 391. 



THE DANGEROUS SECTION OF NON-PRIS- 
MATIC BEAMS. 

277. Remarks. By " dangerous section " is meant that sec- 
tion (in a given beam under given loading with given mode 
of support) where p, the normal stress in the outer fibre, 
at distance e from its neutral axis, is greater than in the 
outer fibre of any other section. Hence the elasticity of 
the material will be first impaired in the outer fibre of 
this section, if the load is gradually increased in amount 
(but not altered in distribution). 

In all preceding problems, the beam being prismatic, /, 
the moment of inertia, and e were the same in all sections, 

hence when the equation P-=3f [§239] was solved for p, 

e 

. . 31 e «* 

giving p=— . . . . (1) 

we found that p was a max., == p m , for that section whose 
31 was a maximum, since p varied as 31, for the moment 



FLEXURE KON-PRISMATIC BEAMS. 



333 



of the stress-couple, as successive sections along the beam 
were examined. 

But for a non-prismatic beam /and e change, from sec- 
tion to section, as well as 31, and the ordinate of the 
moment diagram no longer shows the variation of p, nor 
is p a max. where M is a max. To find the dangerous 
section, then, for a non-prismatic beam, we express the M y 
the 7, and the e of any section in terms of x, thus obtain- 
ing £>=func. (x), then writing dp-r-dx=0, and solving for x. 

278. Dangerous Section in a Double Truncated Wedge. Two 
End Supports. Single Load in Middle. — The form is shown in 
Fig. 284. Neglect weight of beam ; measure x from one sup- 




port 0. The 
\ reaction at 



Fig. 284. 



■c — >\ 



,_ P. The 
width of the 
beam = b at 

all sections, while its height, v, varies, being = h at 0. 
To express the e = y 2 v, and the I = 1 bv z (§247) of any 
section on OC, in terms of x, conceive the sloping faces 
of the truncated wedge to be prolonged to their intersec- 
tion A, at a known distance = c from the face at 0. We 



then have from similar triangles 
v : x -f- c : : h : c, .*. v 



~(x + c) 

c 



(1) 



and .:e = y 2 - (x+c) and J=i& -^-fc] 3 . (2) 



For the free body nO, I (moms. n ) = gives 

e I! 



(3) 



[That is, the M = y 2 Px.] But from (2), (3) becomes 
p=3P ± . X • an d d P- 3P * (x+cf-2x(x+c ) 

By putting dp -4- dx = we obtain both x = — e, and 



334 



MECHANICS OF ENGINEERING. 



x = + c, of which the latter, x = + c, corresponds to a 
maximum for p (since it will be found to give a negative 
result on substitution in d 2 p -=- dx 2 ). 

Hence the dangerous section is as far from the support 
0, as the imaginary edge, A, of the completed wedge, but 
of course on the opposite side. This supposes that the 
half -span, y 2 l, is > c ; if not, the dangerous section will 
be at the middle of the beam, as if the beam were 
prismatic. 



Hence, with ) ^equation for safej BW 

wj' c V loading is: (A'=heighU — ^— =%Pl 



/ 2t < c ) at middle) 

while with ) the equation for safe 
T/7 . V loading is: (put x—q 

y° l > c jandj>=B'i£[3]) 

(see §239.) 






(5) 



MfV-fiPo (6) 



279. Double Truncated Pyramid and Cone. Fig. 285. For 




Fig. 285. 



the truncated pyramid both width = u, and height = v, 
are variable, and if b and h are the dimensions at 0, and 
c = ~0A — distance from to the imaginary vertex A, we 

shall have from similar triangles u=^ (a?+c)and v=-(x+c). 

c c 

Hence, substituting e=j4v and I=~ uv z , in the moment 
equation 



pi Px , c 8 07 
Tr=0, we have^>=3P_ - . 3 

# 'cfaj M 2 ' (z+c) 6 



(7) 

(8) 



FLEXURE. NCXN-PBISMATIC BEAMS. 335 

Putting this = 0, we have x = — c, x = — c, and x = 
+ y 2 c, hence the dangerous section is at a distance x = y 2 c 
from 0, and the equation for safe loading is 

either ^-^% PI . . . . if % I is < % c .... (9) 

(in which V and ¥ are the dimensions at mid-span) 

or MIlMMl = % Pa if y 2 i is > y 2 . . . (10 ) 

For the truncated cone (see Fig. 285 also, on right) where 
e = the variable radius r, and I = J^ >r r 4 , we also have 

' =[Const] • d~c f oi) 

and hence p is a max. for x = ^ c, and the equation for 
safe loading 

either 5fL!L = # PZ, for # Z < ^ c . . . . . (12) 
(where r' = radius of mid-span section) ; 

or *B> (|r ) 3 = ly/ pc f Qr lA l > , ^ c 

4 
(where r = radius of extremity.) 



NON-PRISMATIC BEAMS OF "UNIFORM 
STRENGTH." 

280. Remarks. A beam is said to be of " uniform 
strength " when its form, its mode of support, and the dis- 
tribution of loading, are such that the normal stress p has 
the same value in all the outer fibres, and thus one ele- 
ment of economy is secured, viz. : that all the outer fibres 
may be made to do full duty, since under the safe loading, 
p will be = to R f in all of them. [Of course, in all cases 
of flexure, the elements between the neutral surface and 



336 



MECHANICS OF ENGINEERING. 



the outer fibres being under tensions and compressions 1 
less than B' per sq. inch, are not doing full duty, as. 
regards economy of material, unless perhaps with respect 
to shearing stresses.] In Fig. 265, §261, we have already 
had an instance of a body of uniform strength in flexure, 
viz. : the middle segment, CD, of that figure ; for the 
moment is the same for all sections of CD [eq. (2) of that 
§], and hence the normal stress p in the outer fibres (the 
beam being prismatic in that instance). 

In the following problems the weight of the beam itself 
is neglected. The general method pursued will be to find 
an expression for the outer-fibre-stress p, at a definite sec- 
tion of the beam, where the dimensions of the section are 
known or assumed, then an expression for p in the varia- 
ble section, and equate the two. For clearness the figures 
are exaggerated, vertically. 

281. Parabolic Working Beam. TJnsymmetrical. Fig. 286 




Fig. 386. 



CBO is a working beam or lever, B being the fixed fulcrum 
or bearing. The force P being given we may compute P c 
from the mom. equation P l = P c l lf while the fulcrum 
reaction is P B =P -}-P c , All the forces are "| to the beam. 
The beam is to have the same width b at all points, and is 
to be rectangular in section. 

Ee quired first, the proper height h x , at B, for safety. 
From the free body BO, of length = l , we have I (momSfi} 



0; i.e., 



PbT_ 



PX J or p B 



6 PI 



• (1) 



FLEXUKE. XON-PBJSMATIC BEAMS. 337 

Hence, putting p B = B', 7^ becomes known from (1). 

Bequired, secondly, the relation between the variable 
height v (at any section n) and the distance x of n from 0. 
For the free body nO, we have (I moms n — 0) 

pJ =FqX . or Pn 1 by* =Px and m . m = QP^ (2) 

But for " uniform strength " p Q must = ^? B ; hence 
equate their values from (1) and (2) and we have 

5 = ^ which may be written {% vf = H^lL x (3) 
v 2 hi 2 l 

so as to make the relation between the abscissa x and the 
ordinate y 2 v more marked ; it is the equation of a para- 
bola, whose vertex is at 0. 

The parabolic outline for the portion BC is found simi- 
larly. The local stresses at C, B, and must be proper- 
ly provided for by evident means. The shear J = P , at 
0, also requires special attention. 

This shape of beam is often adopted in practice for the 
working beams of engines, etc. 

The parabolic outlines just found may be replaced by 
trapezoidal forms, Fig. 287, without using much more ma- 
terial, and by making the slop- 
ing plane faces tangent to the 
parabolic outline at points T 
and T 1} half-way between and 
B, and C and B, respectively. It 
can be proved that they contain minimum volumes, among 
all trapezoidal forms capable of circumscribing the given 
parabolic bodies. The dangerous sections of these trape- 
zoidal bodies are at the tangent points T and T lt This is 
as it should be, (see § 278), remembering that the subtan- 
gent of a parabola is bisected by the vertex. 

282. I-Beam of Uniform Strength. — Support and load same 
as in the preceding §. Fig. 288. Let the area of the 




338 



MECHANICS OF ENGINEERING. 




Fig. 288. 

flange -sections be = F and let it be the same for all values 
of x. Considering all points of F at any one section as at 
the same distance z from the neutral axis, we may write 
/ — z 2 F, and assuming that the flanges take all the tension 
and compression while the (thin) web carries the shear, the 
free body of length x in Fig. 288 gives (moms, about n) 



pl_ 



*-=P c x; i.e. 

e 



vz 2 F 
JL —P c x : or, since p is to be constant, 



z = [Const.], x (1) 

i. e. z must be made proportional to x. 

Hence the flanges should be made straight. Practically, 
if they unite at C, the web takes but little shear. 



283. Rectang. Section. Height Constant. Two Supports (at Ex- 
tremities). Single Eccentric Load. 
— Fig. 289. b and h are the 
dimensions of the section at 
B. With BO free we have 

P^-P o l =Or.p B =^ (1) 
e B oh? 

At any other section on BO, as n, where the width is u f 
the variable whose relation to x is required, we have for 
nOiree 




PnL 



-.P x;ov^{^ J l=P x .'. Pt 



6P () x 



e v uh 2 

Equating p B and p n we have u :b :: x :l 



(2) 
(3) 



FLEXURE. BEAMS OF UNIFORM STRENGTH. 339 

That is, BO must be wedge-shaped with its edge at 0, ver- 
tical. 

284. Similar Rectang. Sections. Otherwise as Before. — Fig. 289 a, 
b and h are the dimen- 
sions at i?; at any other 
section n, on BO, the height 
v and width u, are the _° 
variables whose relation to 
■ x is desired and by hypoth- 
esis are connected by the 
relation 




Fig. 289 a. 



u : v :: b : h 



(i) 



(since the section at n is a rectangle similar to that at B). 

• • • (2) 



For the free body BO p B = 6 ^f> 

Oft 



For the free body nO 



Pn 



_6PpX 

uv 2 



(3) 



Writing p a == p B we obtain l -r- bh 2 = x -f- uv 2 , in which 
put u = bv -i- h, from (1) ; whence 



; or (y 2 vY=(y 2 hf: 



(4) 



which is the equation of the curve (a cubic parabola) 
whose abscissa is x and ordinate ]/ 2 v ; i.e., of the upper 
curve of the outline of the central longitudinal vertical 
plane section of the body (dotted line BO) which is sup- 
posed symmetrical about such a plane. Similarly the 
central horizontal plane section will cut out a curve a 
quarter of which (dotted line B* 0) has an equation 



y 2U f={y 2 by* 



(5) 



That is, the height and width must vary as the cube root 



340 MECHANICS OF ENGINEERING. 

of the distance from the support. The portion CB will 
give corresponding results, referred to the support G. 

[If the beam in this problem is to have circular cross- 
sections, let the student treat it in the same manner.] 

286. Uniform Load. Two End Supports. Rectangular Cr. 

Sections. Width Constant. — j^ ^ z ^j- w r 

Weight of beam neglected. * P ^ ^1 I 

How should the height vary, j^^ ~f ~^-~^H^7| 
the height and width at the pT^pi — -r \ '^"^""Pi? 
middle being Ji and b ? Fig. LI Y \ \ J, B '| I I \ \ \(/ 
289 5. From symmetry each '— ^ . _ , I ' 

reaction = j4 W = j4 wl. Fie. 289 ». 

At any cross section n, the width is = b, (same as that at 
the middle) and the height == v, variable. I (moms. n ) =0, 
for the free body nO, gives 

&^= #«fc- ffiia*^ = */ 2 wlx- "£ ... (1) 

e n 2 ' %v 2 

But for a beam of uniform strength, p a is to be = p B as 
computed from 2 (moms. B ) = for the free body . . BO, 
i.e from 

PBVJ» =}iv f .1 _H Z M . . . (2 ) 

feh 2 2 

Hence solve (1) for p n and (2) for p B and equate the results, 

whence tf= # t [h^*] ; or (y 2 vf=i^[Ix-^ . (3) 

This relation between the abscissa x and the ordinate y^v, 
of the curve CBO, shows it to be an ellipse since eq. (3) 
is that of an ellipse referred to its principal diameter and 
the tangent at its vertex as co-ordinate axes. 

In this case eq. (3) covers the whole extent of both 
upper and lower curves, i.e. the complete outline, of the 
curve CBOB', whereas in Figs. 286, 289, and 289 a, such is 
not the case. 



FLEXUKE. BODIES OF UNIFORM STRENGTH. 341 

287. Cantilevers of Uniform Strength. — Beams built in at 
one end, horizontally, and projecting from the wall with- 
out support at the other, should have the forms given be- 
low, for the given cases of loading, if all cross-sections are 
to be Rectangular and the weight of beam neglected. Sides 
of sections horizontal and vertical. Also, the sections are 
symmetrical about the axis of the piece, b and h are the 
dimensions at the wall. 1= length. No proofs given. 




Fig. 290. 




Width constant. -^ 
Vertical outline I 
parabolic. Single \ Fl & 290 > ^ 
end load. 



«x 



MvfH 1 /^ (1) 



Height constant. 
Single end load. 
Horizontal outline 
triangular. 



x 



Fig. 290, (b). (y 2 u)=(y 2 b)* . (2) 



Constant ratio of 
height v to width u. 



{%vf^{y 2 hf% 



Both outlines m - \ Fi * 29 °- ^ ^y^y* 



bic parabolas. 



i 



(3) 
(3Y 



342 MECHANICS OF ENGINEERING. 

Uniform Load.") 
"Width constant. | _. ^ M , , , * V , „s x 
Vertical outlinetri- N' 291 > (<* W»5=W») 7 • (4) 
angular. J 



Uniform Load. 
Height constant. 
Horiz. outline is 
two parabolas meet- 
ing at (vertex) 
with geomet. axes 
II to wall. 



Fig. 291,(6), #«'=(#&)? - .(5) 



Uniform Load.-j a? 2 

Both outlines semi- | (# w )*=(#&)' 7 (6) 

cubic parabolas. V Fig. 291, (c). 



Sections similar 
rectangles. 



*)•=(#*)' TT (6)' 



289. — Beams and cantilevers of circular cross-sections 
may be dealt with similarly, and the proper longitudinal 
outline given, to constitute them " bodies of uniform 
strength." As a consequence of the possession of this 
property, with loading and mode of support of specified 
character, the following may be stated ; that to find the 
equation of safe loading any cross-section tvhatever may be 
employed. This refers to tension and compression. As 
regards the shearing stresses in different parts of the beam 
the condition of " uniform strength " is not necessarily ob- 
tained at the same time with that for normal stress in the 
outer fibres. 



DEFLECTION OF BEAMS OF UNIFORM 
STRENGTH. 

290. Case of § 283, the double wedge, but symmetrical, 
i.e., li=l =^l, Fig. 292. Here we shall find the use of the 



BEAMS OF UNIFORM STRENGTH. 



343 




Fig. 292. 



EI 



form — (of the three forms for the moment of the stress 

couple, see eqs. (5), (6) and (7), §§ 229 and 231) of the most 
direct service in determining the form of the elastic curve 
OB, which is symmetrical, and has a common tangent at 
B, with the curve BG. First to find the radius of curva- 
ture, p, at any section n, we have for the free body n0 9 
2 7 (moms. n =0), whence 

EI ,-n r, i , f from eq. ) u ., ,,.,,,- 

—— + I /£Px=0 ; but J ( 3 ) § 283 J x= h^ and I== l^ uJl 



we have 1 / 1 



— uh 6 = %Jr -T and 



P-K T 



W 
I 



E_ 
P 



a) 



from which all variables have disappeared in the right 
hand member ; i.e., p is constant, the same at all points of 
the elastic curve, hence the latter is the arc of a circle, 
having a horizontal tangent at B a 

To find the deflection, d, at B, consider Fig. 292, (b\ 
where d=KB, and the full circle of radius BB=p i$ 
drawn. 

The triangle KOB is similar to YOB, 
and .\ KB lOB :~. OB : YB 
But 0B=j4l, KB=d and YB=2p 



d= 



¥ P 



and o*. , from eq. (1), d=}i. 



WE 



m 



From eq. (4) §233 we note that for a beam of the same 
material but prismatic (parallelopipedical in this case,) 
having the same dimensions, b and h, at all sections as at 

1 Pf , PV 
WE 



the middle, deflects an amount ~jo 'wi =1 Ap^ un( ^ er a 



344 



MECHANICS OF ENGINEERING. 



load P in the middle of the span. Hence the tapering 
beam of the present § has only 2 /$ the stiffness of the pris- 
matic beam, for the same b, h, I, U, and P. 



291. Case of §281 (Parabolic Body), With^=? , i.e., Symmet- 
rical. — Fig. 293,(a). Required the equation of the neutral 




Fig. 293. 



line OB. For the free body nO, 2'(moms. n )=0 gives us 



*•&—**• 



(i)' 



Fig. 293, (5), shows simply the geometrical relations of the 
problem, position of origin, axes, etc. OnB is the neutral 
line or elastic curvp whose equation, and greatest ordinate 
d, are required. (The right hand member of eq. (1)" is made 
negative because d 2 y-k-dx 2 is negative, the curve being con- 
cave to the axis Xin this, the first quadrant.) 

Now if the beam were prismatic, i", the " moment of in- 
ertia " of the cross-section would be constant, i.e., the same 
for all values of x, and we might proceed by taking the x- 
anti-derivative of each member of (1)" and add a constant ; 
but it is variable and is 



=!&z/3=JL.. 



bh{ A 
hence (1)" becomes 

12 



(fromeq. 3, §281, putting l =j4l) 



WW 



x* 



dx 2 A 



Px 



(1)' 



*y. 



To put this into the form Const. X -=^=func. of (x), we need 



BEAMS OF UNIFORM STRENGTH. 345 



3 

only divide through by x 2 ,, (and for brevity denote 

3 

A Ebh\-r- (%l)Y by A) and obtain 

We can now take the cc-anti-derivative of each member, and 
have 

A^—iAPCZx+y^+C .... (2)' 
ax 

To determine the constant C, we utilize the fact that at B, 
where x=*4l } the slope dy-r-dx is zero, since the tangent 
line is there horizontal, whence from (2)' 

0=-P V J+G .:C=Pj± 
.-.(2)' becomes^ $L =P[</}£l-x #] (2) 



dx 



3 



^=P[V]f^-^^]+[(7'=0] ... (3) 

((7'=0 since for x=0, y=0). We may now find the deflec- 
tion d (Fig. 293(5)) by writing x=j4lsnid y=d } whence, after 
restoring the value of the constant A, 

*-*»£ w 

p73 

and is twice as great [being=2. .]* as if the girder 

4.A 6/^ 

i 3 

* See § 233, putting / =^ bh in e q. (4). 

were parallelopipedical. In other words, the present girder 
is only half as stiff as the prismatic one. 

292. Special Problem. (I.) The symmetrical beam in Fig. 
294 is of rectangular cross-section and constant width = b, 



346 



MECHANICS OF ENGINEERING. 



but the height is constant only over the extreme quarter 
spans, being —li\= x /2 K i. e., half the height h at mid-span. 
The convergence of the two truncated wedges forming the 
middle quarters of the beam is such that the prolongations 




Fig. 294. 



of the upper and lower surfaces would meet over the supports 
(as should be the case to make h=2h i ). Neglecting the 
weight of the beam, and placing a single load in middle, it 
is required to find the equation for safe loading ; also the 
equations of the four elastic curves ; and finally the deflec- 
tion. 

The solutions of this and the following problem are left 
to the student, as exercises. Of course the beam here 
given is not one of uniform strength. 

293. Special Problem. (II). Fig. 295. Eequired the man- 
ner in which the width of the beam must vary, the height 
being constant, cross-sections rectangular, weight of beam 




Fig. 295. 



neglected, to be a beam of uniform strength, if the load is 
uniformly distributed ? 



FLEXURE. OBLIQUE FORCES. 347 



CHAPTER V. 

FLEXURE OF PRISMATIC BEAMS UNDER 
OBLIQUE FORCES. 



294 Remarks, By " oblique forces " will be understood 
external forces not perpendicular to tho beam, but these 
external forces will be confined to one plane, called the 
force-plane, which contains the axis of the beam and also 
cuts the beam symmetrically. The curvature induced by 
these external forces will as before be considered very 
slight, so that distances measured along the beam will be 
treated as unchanged by the flexure. 

It will be remembered that in previous problems the 
proof that the neutral axis of each cross section passes 
through its centre of gravity, rested on the fact that when 
a portion of the beam having a given section as one of its 
bounding surfaces is considered free, the condition of 
equilibrium I (compons. || to beam)=0 does not introduce 
any of the external forces, since these in the problems re- 
ferred to, were "1 to the beam ; but in the problems of the 
present chapter such is not the case, and hence the neutral 
axis does not necessarily pass through the centre of gravity 
of any section, and in fact may have only an ideal, geomet- 
rical existence, being sometimes entirely outside of the 
section ; in other words, the fibres whose ends are exposed 
in a given section may all be in tension, (or all in compres- 
sion,) of intensities varying with the distance of each from 
the neutral axis. It is much more convenient, however, to 
take for an axis of moments the gravity axis parallel to the 



348 



MECHANICS OF ENGINEERING. 



neutral axis instead of the neutral axis itself, since this 
gravity axis has always a known position. 

295. Classification of the Elastic Forces. Shear, Thrust, and 
Stress-Couple. Fig. 296. Let AKM be one extremity of a 
portion, considered free, of a prismatic beam, under oblique 
forces. C is the centre of gravity of the section ex- 
posed, and GG the gravity axis ~\ to the force plane CAK. 
The stresses acting on the elements of area (each —dF ) of 
the section consist of shears (whose sum=J, the "total 
shear") in the plane of the section and parallel to the force 
plane, and of normal stress parallel to AK and proportion- 
al per unit of area to the distances of the dF's on which 
they act from the neutral axis NC", real or ideal (ideal in 
this figure). Imagine the outermost fibre KA, whose dis- 
tance from the gravity axis is=e and from the neutral axis 




/! 

/ i 
/ i 



y / ^ y 









Fig. 296. 



=c4-a, to be prolonged an amount AA\ whose length by 
some arbitrary scale represents the normal stress (tension 
oi compression) to which the dF at A is subjected. Then, 
ii a plane be passed through A' and the neutral axis NC", 
tne lengths, such as mr, parallel to AA', intercepted between 
this plane and the section itself, represent the stress-inten- 



FLEXURE. OBLIQUE FORCES. 349 

sities (i. e., per unit area) on the respective dF's. (In this 
particular figure these stresses are all of one kind, all ten- 
sion or all compression ; but if the neutral axis occurs 
within the limits of the section, they will be of opposite 
kinds on the two sides of NC") Through C, the point 
determined in A'NC" by the intercept CC of the centre of 
gravity, pass a plane A"M"T" parallel to the section it- 
self ; it will divide the stress -intensity AA' into two parts 
Pi and p 2 , and will enable us to express the stress-intensity 
mr, on any dF at a distance g from the gravity -axis GC, in 
two parts ; one part the same for all dF's, the other depen- 
dent on z, thus : 

[Stress-intensity on any dF~\ = pi+ —p 2 . . (1) 

e 

and hence the 

z 
[actual normal stress on any dF] = p x d F + — . p 2 dF (2) 

For example, the stress-intensity on the fibre at T, where 

p 
z = — e l} will be p x — — p 2 , and it is now seen Low we may 

find the stress at any dF when p Y and p 2 have been found. 
If the distance a, between the neutral and gravity axes is 
desired, we have, by similar triangles 

p 2 : e :: C C : a whence a =^ . e . . . (3) 

It is now readily seen, graphically, that the stresses or elas- 
tic forces represented by the equal intercepts between the 
parallel planes AMT and A" M" T", constitute a uniform- 
ly distributed normal stress, which will be called the " uni- 
form thrust," or simply the thrust (or pull, as the case may 

be) of an intensity = p l9 and .*. of an amount = Cp^dF = 

Vi fdF = p,F. 

It is also evident that the positive intercepts forming the 



350 



MECHANICS OF ENGINEERING. 



wedge A"A'G'C f and the negative intercepts forming the 
wedge WM'G'C' form a system of "graded stresses" 
whose combination (algebraic) with those of the " thrust " 
shows the two sets of normal stresses to be equivalent to 
the actual system of normal stresses represented by the 
small prisms forming the imaginary solid AMT . . A'WT'. 
It will be shown that these graded stresses constitute a 
" stress-couple." 

Analytically, the object of this classification of the nor- 
mal stresses into a thrust and a stress-couple, may be made 
apparent as follows : 

In dealing with the free body KAM Fig. 296, we shall 
have occasion to sum the components, parallel to the beam, 
of all forces acting (external 
and elastic), also those ~| to 
the beam; and also sum their 
moments about some axis "| 
to the force plane. Let this 
axis of moments he GO the 
gravity -axis of the section 
(and not the neutral axis) ; 
also take the axis X || to the 
beam and F"| to it (and in r\ , 

force-plane). Let us see \j 

what part the elastic forces FlG . 297 . 

will play in these three summations. See Fig. 297, which 
gives merely a side view. Eeferring to eq. (2) we see that 



p 2 dF-\ 




elastic"! 



[The.^ofthe— ™J=^ dJ?+*jT 



zdF 



i.e. 



i( t( 



-p.F +'-^2 



Si 
e 



[see eq. (4) § 23]. But as the z's are measured from G a 
gravity axis, z must be zero. Hence 

[The IX of the Elastic forces] =PiF= j **£ t ^ \ (4) 



FLEXURE. OBLIQUE FOECES. 351 

Also, 

[The lYoi the Elastic forces] = J= the shear; . (5) 
while for moments about G [see eq. (1)] 
[The 2 (moms. G ) of the elastic forces] = 

f% Pl dF)z+fX^p 2 dF)z 
=p i r'zdF+^f°z 2 dF 

\ \ 

and hence finally 

{i^i^7oii}=^ G = ^^ • • • w 

e 

where I G , — j z 2 dF, is the "moment of inertia " of 
i 
the section about the gravity axis G, (not the neutral axis). 
The expression in (6) may be called the moment of the 
stress-couple, understanding by stress-couple a couple to 
which the graded stresses of Fig. 297 are equivalent. That 
these graded stresses are equivalent to a couple is shown 
by the fact that although they are X forces they do not 
appear in eq. 4, for IX\ hence the sum of the tensions 

^~ C zdF equals that of the compressions ?1 C zdF 

in that set of normal stresses. 

We have therefore gained these advantages, that, of the 
three quantities J" (lbs.), p l (lbs. per sq. inch), and p 2 (lbs. 
per sq. inch) a knowledge of which, with the form of the 
section, completely determines the stresses in the section, 
equations (4), (5), and (6) contain only one each, and hence 
algebraic elimination is unnecessary for finding any one 
of them ; and that the axis of reference of the moment of 
inertia 7 is the same axis of the section as was used in 
former problems in flexure. 

Another mode of stating eqs. (4), (5) and (6) is this : The 
sum of the components, parallel to the beam, of the exter- 
nal forces is balanced by the thrust or pull; those perpen- 



352 MECHANICS OF ENGINEERING. 

dicular to the beam are balanced by the shear; while the 
sum of the moments of the external forces about the 
gravity axis of the section is balanced by that of the stress- 
couple. Notice that the thrust can have nojnoment about 
the gravity axis referred to. 

The Equation for Safe Loading, then, will be this : 



(a) . Oi±j9 2 ) max. 



or 



( h ) • (P^jPi) max. 



whichever 

is 
greater. 



B> , . (7) 



For R', see table in § 251. The double sign provides for 
the cases where p l and p 2 are of opposite kinds, one tension 
the other compression. Of course (pi+i^) max is not the 
same thing as \_p Y max. + p 2 max.]. Inmost cases in prac- 
tice e!=e, and then the part (b) of eq. (7) is unnecessary. 

295a. Elastic Curve with Oblique Forces. — (By elastic curve 
is now meant the locus of the centres of gravity of the sec- 
tions.) Since the normal stresses in a section differ from 
those occuring under perpendicular forces only in the ad- 
dition of a uniform thrust (or pull), whose effect on the 
short lengths (=dx) of fibres between two consecutive sec- 
tions TFV and U V , Fig. 297, is felt equally by all, the loca- 
tion of the centre of curvature R, is not appreciably differ- 
ent from what it would be as determined by the stress- 
couple alone. 

Thus (within the elastic limit), strains being proportional 
to the stresses producing them, if the forces of the stress- 
couple acted alone, the length dx— G G' of a small portion 
of a fibre at the gravity axis would remain unchanged, and 
the lengthening and shortening of the other fibre-lengths 
between the two sections Z7 V and T7 V\ originally parallel, 
would occasion the turning of U' V through a small angle 
(relatively to U V ) about G\ into the position which it oc- 
cupies in the figure (297), and G R l would be the radius of 
curvature. But the effect of the uniform pull (added to 
that of the couple) is to shift U' V parallel to itself into 
the position UV, and hence the radius of curvature of the 






FLEXURE. OBLIQUE FORCES. 



353 



elastic curve, of which G G is an element, is G B instead 
of G B'. But the difference between G Q B and G Q B' is very 
small, being the same, relatively, as the difference between 
G Q G and G G' ; for instance, with wrought-iron, even if p i9 
the intensity of the uniform pull, were as high as 22,000 
lbs. per sq. in. [see § 203] G Q G would exceed G G' by only 
y i2 of one per cent. (=0.0008) ; hence by using GB' instead 
of GB as the radius of curvature p, an error is introduced 
of so small an amount as to be neglected. 



But from § 231, eqs. (6) and (7), 



EI 

P 



= Ei*y 

dx 2 



M, the 



the sum of the moments of the external forces ; hence for 
prismatic beams under oblique forces we may still use 



±EI% { =M) 



(1) 



as one form for the -(moms.) of the elastic forces of the 
section about the gravity-axis ; remembering that the axis 
X must be taken parallel to beam. 

296. Oblique Cantilever with Terminal Load. — Fig. 298. Let 
1= length. The "fixing" of the lower end of the beam is 
its only support. Measure x along the beam from 0. Let 




Fig. 298. 



Fig. 299. 



n be the gravity axis of any section and nT, =x sin «, the 
length of the perpendicular let fall from n on the line of 
action of the force P (load). The flexure is so slight that 
nT is considered to be the same as before the load is al- 



354 MECHANICS OF ENGINEERING. 

lowed to act. [If a were very small, however, it is evident 
that this assumption would be inadmissible, since then a 
large proportion oinT would be due to the flexure caused 
by the load.] 

Consider nO free, Fig. 299. In accordance with the pre- 
ceding paragraph (see eqs. (4), (5), and (6)) the elastic 
forces of the section consist of a shear J, whose value may 
be obtained by writing 2T=0 

whence J=P sin a ; . . . . (1) 

of a uniform thrust =piF, obtained from 2X=0, viz : 

P cos a—p 1 F=0 .'. p,F=P cos a ; . (2) 

and of a stress-couple whose moment [which we may write 

either ^-, or EI C ^ 1 is determined from 2Ymoms. n )=0 or 

e dx 2 

PJ—Px sin a=0, or &I=Px sin a . . (3) 

e e 

As to the strength of the beam, we note that the stress-in- 
tensity, p lt of the thrust is the same in all sections, from 
to L (Fig. 298), and that p 2 , the stress-intensity in the outer 
fibre, (and this is compression if e=no' of Fig. 299) due to 
the stress-couple is proportional to x ; hence the max. of 
[P1+P2] will be in the lower outer fibre at L } Fig. 298, 
where x is as great as possible, =1 ; and will be a compres- 
sion, viz. : 

[jh-Hft] max.=P |-cy + Z(siy) g] . . (4) 

.\ the equation for Safe Loading is 

a, = P r cos« + ?(sin«)e -| # a # (5) 

since with e^e, as will be assumed here,[jo — — p 2 ] max. 



FLEXURE. OBLIQUE FORCES. 



355 



can not exceed, numerically, [P1+P2] max. The stress- 
intensity in the outer fibres along the upper edge of the 
beam, being =p l — p 2 (supposing e x =e) will be compressive 
at the upper end near 0, since there p 2 is small, x being 
small ; but lower down as x grows larger, p 2 increasing, a 
section may be found (before reaching the point L) where 
p 2 =Pi and where consequently the stress in the outer fibre 
is zero, or in other words the neutral axis of that section 
passes through the outer fibre. In any section above that 
section the neutral axis is imaginary, i.e., is altogether out- 
side the section, while below it, it is within the section, but 
cannot pass beyond the gravity axis. Thus in Fig. 300, O'U 




Fig. 



Fig. 301. 



is the locus of the positions of the neutral axis for successive 
sections, while OL the axis of the beam is the locus of the 
gravity axes (or rather of the centres of gravity) of the 
sections, this latter line forming the " elastic curve " un- 
der flexure. As already stated, however, the flexure is to 
be but slight, and a must not be very small. For in- 
stance, if the deflection of from its position before flex- 
ure is of such an amount as to cause the lever-arm OB of 
P about L to be greater by 10 per cent, than its value 
(=1 sin a) before flexure, the value of p 2 as computed from 
eq. (3) (with x=l) will be less than its true value in the 
same proportion. 

The deflection of from the tangent at L, by § 237, Fig. 
229(a) is d=(P sina)P-±-3EI, approximately, putting P sin a 



356 



MECHANICS OF EXGIXEEEIXG. 



for the P of Fig. 229 ; but this very deflection gives to the 
other component, P cos a, || to the tangent at Z, a lever 
arm, and consequent moment, about the gravity axes of all 
the sections, whence for I (moms. L )=0 we have, (more ex- 
actly than from eq. (3) when x=l) 



&L= P (sin a)l+ P eoaa .&™B 



3 EI 



(6) 



(We have supposed P replaced by its components || and 
"| to the fixed tangent at L, see Fig. 301). But even (6) 
will not give an exact value for p 2 at L ; for the lever arm 
of P cos a, viz. d, is >(P sina)? 3 -^3-£T, on account of the 
presence and leverage of P cos a itself. The true value of 
d in this case may be obtained by a method similar to that 
indicated in the next paragraph. 

297. Elastic Curve of Oblique Cantilever with Terminal Load. 
More Exact Solution. For variety place the cantilever as in 

Fig. 302, so that the deflection 
OY=d tends to decrease the 
moment of P about the gravity 
axis of any section, n. We 
may replace P by its X and 7 
components, Fig. 303, || and 
"1 respectively to the fixed 
tangent line at L. The origin, 
0, is taken at the free end of 
the beam. Let a= angle bet- 




Fig. 303. 



Fig. 302. 

ween P and X. For a free body On, n being any section, 
we have I (moms. n )=0 



whence 



EL 



dx 2 



P(cos a)y — P(sin a) x 



(i) 



[See eq. (1) § 295a]. In this equation the right hand 
member is evidently (see fig. 303) a negative quantity; 
this is as it should be, for Eldu-y-^x 2 is negative, the curve 
being concave to the axis X in the first quadrant. (It 
must be noted that the axis X is always to be taken || to 
the beam, for EWy-^dx 2 to represent the moment of the 
stress-couple.) 



FLEXUEE. OBLIQUE FORCES. 357 

Eq. (1) is not in proper form for taking the a>anti-deri- 
vative of both members, since one term contains the vari- 
able y, an unknown function of x. Its integration is in- 
cluded in a more general case given in some works on cal- 
culus, but a special solution by Prof. Robinson, of Ohio, 
is here subjoined for present needs.* 

We thus obtain as the equation of the elastic curve in 
Fig. 303, 



/Pcosa r^+e^«T( s ina)a>-^cosa)y1=sinaref--^ x "|. (2) 

In which e n denotes the Naperian Base= 2.71828, an abstract 
number, and q for brevity stands for VPcosa+UI. 

To find the deflection d. we make x—l in (2), and solve 
for y; the result is d. 

The uniform thrust at L is piF=Pcosa .... (3) 
while the stress intensity p 2 in the outer fibre at L, is ob- 



Denoting Pcoa +EI by # s and Psin a+ZJIbj jt? 2 , eq. (1) becomesy~-=g 2 y— p*x . . (6) 



Differentiate (6) then -—^q" 1 -— —P*- Differentiate again : whence ^-=« a — ^- . 
ax 6 ax fj.nf* * /-;-r>2 



<r) 



Letting % 



=u . (8) so that u=q*v-f*c, from (6) we have ^-=4^-and ^=^? 

J dx 3 dx dx* dx* 

See (7) ^- g i Ui w Mch, mult, by 2 du, gives -^ 2 du du=2q*udu 

• '■ f dx constant J, ^-j / 2 du d ? u=2q i J udu+C .\~j^=4HP+C. whence 
\ J 'V^t^ 1 ' 6 '' XJ « l0g e[ "V^l ]— | ftV C.or.xmtang 



dec 

5 



e = Nap. base, /*=-£, H- \/« a +ff 1; 
n n c L ff 2 J 



ore , 

; or C'e — u- *Jv?+C_ . . (9) 
n o-a 



a 2^z ga; /* ox Ce„ 



Square each side of (9); then C'e —2C'e w-f« 2 =w 2 -f--5 ; .*. u= l A C'e 

n n 2 n 2C'q°- 

—qx ) 
qx Ce n f which gives y .. .., .. ., 
•- (see eqs. 6 and 8) ? 2 2/-j3 2 a:=K C'e n — _ _5. f as func. x. Consolidating the 



358 MECHANICS OF ENGINEERING. 

tained from the moment equation for the free body L 

viz: &L=P(sma)1r—P(cosa)d (4) 

in which e— distance of outer fibre from the gravity axis. 
The equation for safe loading is written out by placing 
the values of p l9 p 2 > and d, as derived from equations (2), 
(3), and (4) in the expression 

p 1 +p 2 =B' 

To solve the resulting equation for P, in case that is the 
unknown quantity, can only be accomplished by successive 
assumptions and approximations, since it occurs trans- 
cendentally. 

In case a horizontal tension-member of a bridge-truss is 
subjected to a longitudinal tension P' (due to its position 
in the truss and the load upon the latter) and at the same 
time receives a vertical pressure =P" at the middle, each 
half will be bent in the same manner as the cantilever in 
Fig. 302 ; y 2 P" corresponding to P sin a, and P' to Pcos a. 

p2 qx —qx 

constant factors we now write y = -^x-{-me n —ne . . the equation required . (10) 



To determine the constants m andn(m= C'-h2q 2 ; n= C^C'q*) we first find dy-i-dx, i.e. 

dv 7) 2 Q x — Q x 

by differentiating (10) -^-—^ me u +Q ne n .... (11) x=0 for y=0 



.'. (10) gives . . . 0=0 -{-me — ne n =0 i.e. m— n=0 .*. m=n . (12) Also for x=l 

dv v 2 & —of\ 

— ^-=0.\ (11) gives 0=—+q \ me n +ne n . . (13); .'. with n=m we have m=n= 

r & -&~\ 

e n ~\-e n . . .(14). The equation of the curve, then, substituting (14) in 



—p^-t-q 



qx —qx 



(10) isy=^x—P- JL — 3 — . (15) .-., Substituting for p and q we 

q q qC — qi 

6 n+«n 

/pcosa r Qi — qi~\ r q x — q x ~\ 

nave, (as in §297) ' EJ e n + e n xsina— ycosa =sina e — <? n 



FLEXURE. OBLIQUE FORCES. 



359 



298. Inclined Beam with Hinge at One End. — Fig. 304. Let 
e = e x . Required the equation for safe loading ; also the 
m3ximum shear, there being but one load, P, and that in 
tL,j middle. The vertical wall being smooth, its reaction, 




H, At is horizontal, while that of the hinge-pin being un- 
known, both in amount and direction, is best replaced by 
its horizontal and vertical components B and V , unknown 
in amount only. Supposing the flexure slight, we find 
these external forces in the same manner as in Prob. 1 § 
37, by considering the whole beam free, and obtain 



H=~ cota ; H also = £. cota ; V = P 



■ (i) 



For any section n between and B, we have, from the 
free body nO, Fig. 305, 



uniform thrust = p x F = H cos a 
and from I (moms. n ) = 0, 



P2I. 

e 



Hx sin a 



. (2) 



(3) 



and the shear = J = H sin a = y 2 p cos a (4) 

The max. (Pi+p 2 ) to be found on OB is .-. close above B t 
where x = y 2 I, and is ' 



360 MECHANICS OF ENGINEERING. 

jETcos a . Hie sin a •, . , D rcot a . le~\ ,^ 

= - — + — = — which = P cos « +_ (b) 

In examining sections on GB let the free body be Cn' 9 
Fig. 306. Then from I (longitud. comps.) = 

(the thrust=) p L F = V sin a + H cos a (6)' 

i.e. p 1 F=P[sin a -\- y 2 cos a cot a] (6) 

while, from 2 1 (moms. n ') = 0, 

P*-= J^sc' cos a — i7 ^' sin a (7V 

e 

i.e. ^i^i^Pcosa^' (7) 

Hence (jpi+jh) f° r sections on CB is greatest when a?' 
is greatest, which is when x f — ]/ 2 I, x' being limited be- 
tween x' — and x' = y 2 I, and is 

(*+*,) max. on GB=P cos a [ tan a+ p cot a +%Jf\ (8) 

which is evidently greater than the max. (P1+P2) on BO \ 
see eq. (5). Hence the equation for safe loading is 

B' = P cos g pa n «+J^cot« + ^j . # . . (9) 

in which R' is the safe normal stress, per square unit, for 
the material. 

The shear, J t anywhere on CB, homl (transverse comp.) 
=0 in Fig. 306, is 

J— y cos a — Hq sin a = *£ P cos a . . (10) 



361 



FLEXURE. OBLIQUE FORCES. 



As showing graphically all the results found, moment, 
thrust, and shear diagrams are drawn 
in Fig. 307, and also a diagram whose 
ordinates represent the variation of 
{pi-\-p2) along the beam. Each ordi- 
nate is placed vertically under the 
gravity axis of the section to which it 
refers. 

299. Numerical Example of the Forego- 
ing. — Fig. 308. Let the beam be of 
wrought iron, the load P = 1,800 lbs., 
hanging from the middle. Cross sec- 
tion rectangular 2 in. by 1 in., the 2 
in. being parallel to the force-plane. 
Required the max. normal stress in 
a,ny outer fibre ; also the max. total 
shear. 

This max. stress -intensity will be in 
the outer fibres in the section just below B and on the 
upper side, according to § 298, and is given by eq. (8) of 
that article ; in which, see Fig. 308, we must substitute 
(inch-pound-second-system) P = 1,800 lbs.; F — 2 sq 
I = V120 2 + 12 2 = 120.6 in.; e = l in., I = £ bh 
biquad. inches ; cot a = JL ; cos a = .0996 




. m.; 

1-2/ 

12—73 
and tan a = 10. 



max. (^+p 2 )=1800x.0996 



io+y 2 



120.6: 



- 1 ] 





..^ 



Fig. 308. Fig. 309. Fig. 310. 

9000 lbs. per sq. inch, very nearly, compression. This is vo 



362 MECHANICS OF ENGINEERING. 

tlie upper outer fibre close under B. In the lower outer fibre 
just under B we have a tension = p 2 ~P\— 7,200 lbs. per 
sq. in. (It is here supposed that the beam is secure against 
yielding sideways.) 

300. Strength of Hooks. — An ordinary hook, see Fig. 309, 
may be treated as follows : The load being == P, if we 
make a horizontal section at AB, whose gravity axis g is 
the one, of all sections, furthest removed from the line of 
action of P, and consider the portion C free, we have the 
shear = J = zero (1) 

the uniform pull = p l F=P . . . (2) 

while the moment of the stress-couple, from I (moms. g ) = 
0, is 

&*=Pa (3) 

e 



For safe loading p^ + p 2 must = B', i.e. 



B'=P 



r 1 ae 



(4) 



It is here assumed that e = e l3 and that the maximum 
ta+J^] occurs a ^ AB. 

301 Crane. — As an exercise let the student investigate the 
strength of a crane, such as is shown in Fig. 310. 



FLEXUBE. LONG COLUMNS. 363 



CHAPTER VI. 



FLEXURE OF " LONG COLUMNS.'' 



302. Definitions. — By " long column " is meant a straight 
beam, usually prismatic, which is acted on by two com- 
pressive forces, one at each extremity, and whose length 
is so great compared with its diameter that it gives way 
(or " fails ") by buckling sideways, i.e. by flexure, instead 
of by crushing or splitting like a short block (see § 200). 
The pillars or columns used in buildings, the compression 
members of bridge-trusses and roofs, the " bents " of a 
trestle work, and the piston-rods and connecting-rods of 
steam-engines, are the principal practical examples of long 
columns. That they should be weaker than short blocks 
of the same material and cross-section is quite evident, but 
their theoretical treatment is much less satisfactory than 
in other cases of flexure, experiment being very largely 
relied on not only to determine the physical constants 
which theory introduces in the formulae referring to them, 
but even to modify the algebraic form of those formulae, 
thus rendering them to a certain extent empirical. 

303. End Conditions. — The strength of a column is largely 
dependent on whether the ends are free to turn, or are 
fixed and thus incapable of turning. The former condi- 
tion is attained by rounding the ends, or providing them 
with hinges or ball-and-socket-joints ; the latter by facing 
off each end to an accurate plane surface, the bearing on 
which it rests being plane also, and incapable of turning. 



364 



MECHANICS OF ENGINEERING. 



In the former condition the column is spoken of as having 
round ends ; Fig. 311, (a) ; in the latter as having fixed euds, 
(or flat bases ; or square ends), Fig. 311, (b). 




Fig. 312. 

Sometimes a column is fixed at one end while the other 
end is not only round but incapable of lateral deviation from 
the tangent line of the other extremity ; this state of end 
conditions is often spoken of as "Pin and Square," Fig. 
311, (c). 

If the rounding of the ends is produced by a hinge or 
" pin joint," Fig. 312, both pins lying in the same plane 
and having immovable bearings at their extremities, the 
column is to be considered as round-ended as regards flex- 
ure in the plane "1 to the pins, but as square-ended as re^ 
gards flexure in the plane containing the axes of the pins. 

The " moment of inertia " of the section of a column will 
be understood to be referred to a gravity axis of the sec- 
tion which is "] to the plane of flexure (and this corres- 
ponds to the " force-plane " spoken of in previous chap- 
ters), or plane of the axis of column when bent. 

303«. Euler's Formula. — Taking the case of a round-ended 
column, Fig. 313 (a), assume the middle of the length as 
an origin, with the axis X tangent to the elastic curve at 
that point. The flexure being slight, we may use the form 
EId?y-T-da? for the moment of the stress-couple in any 



FLEXURE. LONG COLUMNS. 



365 




dy 

dx — -V 


dy dy 


dx—-\j 
dx Y7 

~y m" 


-a-y- i 


/ la: 


J 

-a — — A 

\ 


1 
i 


Y 



Pig. 



Fig. 314. 



section n, remembering that with this notation the axis X 
must be || to the beam, as in the figure (313). Considering 
the free body nC, Fig. 313 (&), we note that the shear is 
zero, that the uniform thrust =P, and that 2 , (moms. n )=0 
gives (a being the deflection at 0) 

Multiplying each side by dy we have 



(1) 



El 



dx 



g dyd 2 y=Pady—Pydy 



(2) 



Since this equation is true for the y, dx, dy, and d?y of any 
element of arc of the elastic curve, we may suppose it 
written out for each element from where y =0, and dy=0, 
up to any element, (where dy—dy and y =y) (see Fig. 314) 
and then write the sum of the left hand members equal to 
that of the right hand members, remembering that, since 
dx is assumed constant, 1-z-dx 2 is a common factor on the 
left. In other words, integrate between and any point 
of the curve, n. That is, 

dy=dy 

%L f[dy]d[dy]=Pa f dy-P f v ydy (3) 

The product dy d 2 y has been written (dy)d(dy), (for d 2 y is 



366 MECHANICS OF ENGINEERING. 

the differential or increment of dy) and is of a form like 
xdx, or ydy. Performing the integration we have 

EI dy 2 r> -py 2 , A , 

which is in a form applicable to any point of the curve, 
and contains the variables x and y and their increments 
dx and dy. In order to separate the variables, solve for dx, 
and we have 

— d(l) 

dx= I EI dy _ QTdx=z I EI. \aJ 

V PV2ay-f Vp , y /yy ' ( 5 > 

d(i) a 

.'. J Q dx= ± V p- J i g/y y 

i.e.,a?=±- v /-p- (vers, sin— x |j . . . (6) 

(6) is the equation of the elastic curve DOG, Fig. 313 (a), 
and contains the deflection a. If P and e& are both given, 
y can be computed for a given #, and vice versa, and thus 
the curve traced out, but we would naturally suppose a to 
depend on P, for in eq. (6) when x= y 2 l, y should =a. Mak- 
ing these substitutions we obtain 

#*= Vy ( vers - sin _1 io °) ; ie -> ^ i= Vy \ (7) 

Since a has vanished from eq. (7) the value for P ob- 
tained from this equation, viz.: 

Po=EI^ .... (8) 

is independent of a, and 

is ,\ to be regarded as that force (at each end of the round- 
ended column in Fig. 313) which will hold the column at 
any small deflection at which it may previously have been 
set. 



FLEXURE. LONG COLUMNS. 



In other words, if the force is less than P no flexure at 
all will be produced, and hence P is sometimes called the 
force producing " incipient flexure." [This is roughly ver- 
ified by exerting a downward pressure with the hand on 
the upper end of the flexible rod (a T-square-blade for in- 
stance) placed vertically on the floor of a room ; the pres- 
sure must reach a definite value before a decided buckling 
takes place, and then a very slight increase of pressure oc- 
casions a large increase of deflection.] 

It is also evident that a force slightly greater than P 
would very largely increase the deflection, thus gaining for 
itself so great a lever arm about the middle section as to 
cause rupture. For this reason eq. (8) may be looked 
upon as giving the Breaking Load of a column with round 
ends, and is called Euler' 's formula. 

Eef erring now to Fig. 311, it will be seen that if the three 
parts into which the flat-ended column is di- 
vided by its two points of inflection A and B 
are considered free, individually, in Fig. 315, 
the forces acting will be as there shown, viz.: 
At the points of inflection there is no stress- 
couple, and no shear, but only a thrust, —Pi 
and hence the portion AB is in the condition 
of a round-ended column. Also, the tangents 
to the elastic curves at and C being pre- 
served vertical by the f rictionless guide-blocks 
and guides (which are introduced here simply 
as a theoretical method of preventing the ends 
from turning, but do not interfere with verti- 
cal freedom) OA is in the same state of flex- 
ure as half of AB and under the same forces. 
Hence the length AB must == one half the 
total length I of the flat-ended column. In 
other words, the breaking load of a round- 
ended column of length =}4l, is the same as 
that of a flat-ended column of length =1 
Hence for the I of eq. (8) write y 2 l and we 
have as the breaking load of a column with 
flat-ends and of length =1. 




*l 



Fig. 315. 



368 MECHANICS OF ENGINEERING. 

P,=4i7/iL 2 .... (9) 

Similar reasoning, applied to the " pin-and-square " 
mode of support (in Fig. 311) where the points of inflec- 
tion are at B, approximately % I from C, and at the 
extremity itself, calls for the substitution of 2 /^ I for I in 
eq. (8), and hence the breaking load of a " pin-and-square " 
column, of length = I, is 

P >=1 EI i • • • ( 10 > 

Comparing eqs. (8), (9), and (10), and calling the value of 
P x (flat-ends) unity, we derive the following statement : 
The breaking loads of a given column are as the numbers 



1 

flat-ends 



9/16 
pin-and-square 



round-ends 



according to the 
mode of support. 



These ratios are approximately verified in practice. 

Euler's Formula [i.e., eq. (8) and those derived from it, 
(9) and (10)] when considered as giving the breaking load 
is peculiar in this respect, that it contains no reference to 
the stress per unit of area necessary to rupture the material 
of the column, but merely assumes that the load producing 
" incipient flexure ", i.e., which produces any bending at 
all, will eventually break the beam because of the greater 
and greater lever arm thus gained for itself. In the canti- 
lever of Fig. 241 the bending of the beam does not sensibly 
affect the lever-arm of the load about the wall-section, but 
with a column, the lever- arm of the load about the mid- 
section is almost entirely due to the deflection produced. 

.304. Example. Euler's formula is only approximately 
verified by experiment. As an example of its use when 
considered as giving the force producing " incipient flex- 
ure " it will now be applied in the case of a steel T-square- 
blade whose ends are free to turn. Hence we use the 
round-end formula eq. (8) of §303, with the modulus of 
elasticity i?= 30,000,000 lbs. per sq. inch. The dimensions; 



FLEXURE. LOXG COLUMNS. 369 

are as follows : the length I = 30 in., thickness = 35 of an 
inch, and width = 2 inches. The moment of inertia, I r 
about a gravity axis of the section || to the width (the 
plane of bending being || to the thickness) is (§247) 

1 1 / 1 \ 3 1 

1 = _ W= _ X 2 x^oj =i62^00 bi( l uacL mches - 

.\ , with 7T = 22 -4- 7, 

t: 2 30,000,000 22 2 1 _ nfto „ 
P -^i p- 162j0 00 " 7 2 ' 900- 2 ' 031bs * 

Experiment showed that the force, a very small addition 
to which caused a large increase of deflection or side-buck- 
ling, was about 2 lbs. 

305. Hodgkinson's Formulae for Columns. — The principal 
practical use of Euler's formula was to furnish a general 
form of expression for breaking load, to Eaton Hodgkin- 
son, who experimented in England in 1840 upon columns 
of iron and timber. 

According to Euler's formula we have for cylindrical 

columns, /being =}£ nr* = — tzcP (§247), 

64 

for flat-ends . . P Y =A En B . ^ 

i.e., proportional to the fourth power of the diameter, and 
inversely as the square of the length. But Hodgkinson's 
experiments gave for wrought-iron cylinders 

Pi = (const.) x 72 — > an d f° r cas * i ron Pi=(const.) X 

Again, for a square column, whose side = b, Euler's for- 
mula would give 

whilo Hodgkinson found for square pillars of wood 
P^Cconst-OXp 



370 MECHANICS OF ENGINEERING. 

Hence in the case of wood these experiments indicated the 
same powers for b and I as Euler's formula, but with a dif- 
ferent constant factor ; while for cast and wrought iron 
the powers differ slightly from those of Euler. 

Hodgkinson's formulae are as follows, and evidently 
not homogeneous ; the prescribed units should .*. be care- 
fully followed, d denotes the diameter of the cylindrical 
columns, b the side of square columns, 1= length. 

( For solid cylindrical cast iron columns, flat-ends ; 

< Breaking load in tons ) ii 1Cw/ 7. • -, x 3,55 . /7 . *,\ r 

\ of 2,240 lbs. each j =44.16 x (d m inches) -,(lmit.) 

( For solid cylindrical wrought iron columns, flat-ends ; 
\ Breaking load in tons ) 10iw//J . • -. x 355 . n • <., x 2 
\ of 2,240 lbs. each [ =134 X (dm inches) + (Z in ft.) 

f For solid square columns of dry oak, flat-ends ; 
-I Breaking load in tons ) inne w /•*• • -u \ . n • *± \ 2 
I of 2,240 lbs. each \ =la9 ° x $ m mohe8 ^ -" (* lnfi ) 

f For solid square columns of dry fir, flat-ends ; 

i^itsSetr } = 7 - 81 >< <* * -^ - * - ft -> 2 

Hodgkinson found that when the mode of support was 
" pin-and-square," the breaking load was about y 2 as 
great ; and when the ends were rounded, about % as great 
as with flat ends. These ratios differ somewhat from the 
theoretical ones mentioned in §303, just after eq. (10.) 

Experiment shows that, strictly speaking, pin ends are 
not equivalent to round ends, but furnish additional 
strength ; for the friction of the pins in their bearings 
hinders the turning of the ends somewhat. As the lengths 
become smaller the value of the breaking load in Hodg- 
kinson's formula increases rapidly, until it becomes larger 
than would be obtained by using the formula for the 
crushing resistance of a short block (§201) viz., FG t i.e., 
the sectional area X the crushing resistance per unit of 
area. 

In such a case the pillar is called a short column, or " short 
block," and the value FG is to be taken as the breaking 



FLEXURE. LOXG COLUMNS. 371 

load. This distinction is necessary in using Hodgkinson's 
i ormulae ; i.e., the breaking load is the smaller of the two 
values, FG and that obtained by Hodgkinson's rule. 

In present practice Hodgkinson's formulae are not often 
used except for hollow cylindrical iron columns, for which 
with d 2 and d Y as the external and internal diameters, we 
have for flat-ends 

Breaking load in tons \ _ r , (c? 2 in in.) 3 - 55 — (d l in in.) 3 - 55 
of 2,240 lbs. each [ -^ onst - X (HuTe^ ~~ 

in which the const. = 44.16 for cast iron, and 134 for 
wrought, while n = 1.7 for cast-iron and = 2 for wrought. 

306. Examples of Hodgkinson's Formulae. — Example 1. lie- 
quired the breaking weight of a wrought -iron pipe used 
as a long column, having a length of 12 feet, an internal 
diameter of 3 in., and an external diameter of 3j^ inches, 
the ends having well fitted flat bases. 

If we had regard simply to the sectional area of metal, 
which is F = 1.22 sq. inches, and treated the column as a 
short block (or short column) we should have for its com- 
pressive load at the elastic limit (see table §203) P"=FC" 
= 1.22 x 24,000=29,280 lbs. and the safe load P 1 may be 
taken at 16,000 lbs. 

But by the last formula of the preceding article we have 

Breaking load in ) 1340x ( 3.25) 355 - 3 355 , -- Q - 
tons of 2,240 lbs. each [ _id4 * UX i^ 15,U ' t0nS 

i.e.= 15.07 X 2240=33,768 lbs. 

Detail, [log. 3.25] x 3.55= 0.511883 x 3.55= 1.817184 ; 

[log. 3.00] X 3.55=0.477,121x3.55=1.693,779 ; 

and the corresponding numbers are 65.6 and 49.4; their 
difference = 16.2, hence 

Br. load in long tons = — —15.072 long tons. 

J. "xt: 

=33,768 lbs. 



372 MECHANICS OF ENGINEERING. 

With a " factor of safety " (see §205) of four, we have, as 
the safe load, P' = 8,442 lbs. This being less than the 
16000 lbs. obtained from the " short block " formula,should 
be adopted. 

If the ends were rounded the safe load would be one- 
third of this i.e., would be 2,814 lbs ; while with pin-and- 
square end-conditions, we should use one-half, or 4,221 lbs. 

Example 2. Required the necessary diameter to be 
given a solid cylindrical cast-iron pillar with flat ends, that 
its safe load may be 13,440 lbs. taking 6 as a factor of 
safety. Let d = the unknown diameter. Using the proper 
formula in § 305, and hence expressing the breaking load, 
which is to be six times the given safe load, in long tons 
we have (the length of column being 16 ft.) 

13440x6 _ 44. 16 (din inches) 355 m 

2240 '~~ 16 L7 * ' ' ' { } 

i.e. \d in inches] 355 =?|^|l 7 (2) 

or log.d= s l s [log. 36+1.7xlog. 16-log. 44.16] . . (3) 
,*, log.c?= 3 -l^.[1.958278] =0.551627 .-. d = 3.56 ins. 

This result is for flat ends. If the ends were rounded, 
we should obtain d = 4.85 inches. 

307. Kankine's Formula for Columns. — The formula of this 
name (some times called Gordon's, in some of its forms) has 
a somewhat more rational basis than Euler's, in that it in- 
troduces the maximum normal stress in the outer fibre and 
is applicable to a column or block of any length, but still 
contains assumptions not strictly borne out in theory, thus 
introducing some co-efficients requiring experimental de- 
termination. It may be developed as follows : 

Since in the flat-ended column in Fig. 315 the middle 
portion AB, between the inflection points A and B, is 
acted on at each end by a thrust = P, not accompanied by 
any shear or stress-couple, it will be simpler to treat thai 



FLEXURE. LOXG COLUMNS. 



373 



portion alone Fig. 316, (a), since the thrust and stress- 
couple induced in the section at 
B, the middle of AB, will be equal 
to those at the flat ends, and G, 
in Fig. 315. Let a denote the de- 
flection of B from the straight line 
AB. Now consider the portion 
AB as a free body in Fig. 316, (b), 
putting in the elastic forces of the 
section at B, which may be clas- 
sified into a uniform thrust — 
p x F, and a stress couple of moment fig. 316. 

294). (The shear is evidently zero, from 





Pi, (see 

e 



I (hor comps.) = 0). Here p l denotes the uniform pres- 
sure (per unit of area), due to the uniform thrust, and p 2 
the pressure or tension (per unit of area), in the elastic 
forces constituting the stress-couple, on the outermost 
element of area, at a distance e from the gravity axis ("| 
to plane of flexure) of the section. F is the total area of 
the section. I is the moment of inertia about the said 
gravity axis, g 

2 (vert, comps.) = gives P = p x F . . (1) 



2 (moms.. 



= gives Pa = 



Pil 



(2) 



For any section, n, between A and B, we would evidently 
have the same^ as at B, but a smaller p 2 , since Py < Pa 
while e, /, and F } do not change, the column being pris- 
matic. Hence the max, (pi-fp 2 ) is on the concave edge at 
B and for safety should be no more than G -s- n, where G 
is the Modulus of Crushing (§ 201) and n is a " factor of 
safety." Solving (1) and (2) for p x andj9 2 , and putting their 
sum = G -5- rir we have 



P Pae_G 
F I ~ n 



(3) 



We might now solve for P and call it the safe load, but it 



374 MECHANICS OF ENGINEEKLN G. 

is customary to present the formula in a form for giving 
the breaking load, the factor of safety being applied after- 
ward. Hence we shall make n — 1, and solve for P, call- 
ing it then the breaking load. Now the deflection a is un- 
known, but may be expressed approximately, as follows, 
in terms of e and I. 

Suppose two columns of lengths = V and V , each 

EI' r> 'I' 
bearing its safe load. Then at the point R, — —^jL-; i,e., 

p e 

E'e' = p f p 2 \ Considering the curve AB as a circular arc 

we have (see § 290) a' = V 2 +■ 32 /, i.e. a' = J^ . l Z ; and 

SAE e 

p " V' 2 
similarly for the other column, a" = J? • — , . If the 

oAE 6 

columns are of the same material E f = E" , and if each is 

bearing its safe load we may assume p 2 ' = P2" nearly, in 

which case the term p 2 " 4- E" = p 2 ' 4- E'\ and we may 

say that the deflection a, under safe load, is proportional 

to (length) 2 4- e, approximately, i. e., that ae = pi 2 , where 

ft is a constant (an abstract number also) dependent on 

experiment and different for different materials, and I the 

full length. We may also write, for convenience, I = FW, 

Jc being the radius of gyration (see § 85). Hence, finally, 

we have from eq. (3) 

Breaking load ) _ p _ FG , A ^ 

for flat ends ) 1_ — 7F .... W 

This is known as Rankine's formula. 

By the same reasoning as in § 303, for a round -ended 
column we substitute 2 I for I ; for a pin-and-square col- 
umn 4 I fo r I ; and .*. obtain 



(5) 



Breaking load ) _ p _ F@ 

for a round-ended column { ° W 

Breaking load for )_r>_ FG ,£> 

a pin-and-square column \ ~~ 2 ~ p" • • • V J 



ELEXUKE. LONG COLUMNS. 



375 



These formulae, (4), (5), and (6), unlike Hodgkinson's, 
are of homogeneous form. Any convenient system of units 
may therefore be used in them. 

Rankine gives the following values for C and /?, to be 
u-secl in these formulae. These are based on Hodgkinson's 
experiments. 





Cast Iron. 


Wr't Iron. 


Timber. 


C in lbs. per sq. in. 


80,000 


36,000 


7,200 


/3 (abstract number) 


1 
6,400 


1 


1 


36,000 


3,000 



If these numerical values of C are used F must be ex- 
pressed in Sq. Inches and P in Pounds. Rankine recom- 
mends 4 as a factor of safety for iron in quiescent struct- 
ures, 5 under moving loads ; 10 for timber. The N. J. 
Iron & Steel Co. use Rankine's formula for their wrought 
iron rolled beams, when used as columns, with a factor of 
safety of 4^. 



308. Examples, Using Rankine's Formula. — Example 1. — 
Take the same data for a wrought iron pipe used as a 
column, as in example 1, § 306 ; i.e., 1—12 ft. =144 inches, 
F=%[ti(3}() 2 —7lS 2 ]=1.227 sq. inches, while ¥ for a nar- 
row circular ring like the present section may be put 
= yi{ls/of (see § 98) sq. inches. With these values, and 
(7=36,000 lbs. per sq. in., and P=^^ (for wrought iron), 
we have from eq. (4), for flat ends, 



p _ 1-227x36,000^ =3074a6 lbs . 



(1) 



1+ 



36,000 i [1.625]' 



This being the breaking load, the safe load may be taken 
= y A or y 6 of 30743.6 lbs., according as the structure of 



376 MECHANICS OF ENGINEERING. 

which the column is a member is quiescent or subject to 
vibration from moving loads. By Hodgkinson's formula 
33,768 lbs. was obtained as a breaking load in this case 
(§ 306). 

For rounded ends we should obtain (eq. 5) 

P o =16,100. lbs., as break, load . (2) 

and for pin-and-square, eq. (6) 

P 2 =24,908. lbs. as break, load . . (3) 

Example 2.— (Same as Example 2, § 306). Eequired by 

Rankine's formula the necessary diameter, d, to be given 

a solid cylindrical cast-iron pillar, 16 ft. in length, with 

rounded ends, that its safe load may be six long tons (i.e., 

of 2,240 lbs. each) taking 6 as a factor of safety. F—-£ , 

while the value of k 2 is thus obtained. From § 247, / for 
a full circle about its diameter = }{7rr i =7zr 2 .}£r 2 ,\ ¥=]^r 2 
=1 /wd 2 . Hence eq. (5) of § 307 becomes. 

Po= X**° n . . . (1) 

1+4/? 



16P 

d 2 



P the breaking load is to be =6x6x2,240 lbs., for cast- 
iron is 80,000 lbs. per sq. inch, while (abstract number) 
= air Solving for d we have the biquadratic equation : 

d ,_ 28x6x6x2,240 ^ 2 _ 28x6x6x2,240xl6 2 xl2 2 x4 
22x80,000 — 22x80,000x400 

whence d 2 = 0.641 (1± 33.92), and taking the upper sign, 
finally, d= a/22.4 =4.73 inches. (By Hodgkinson's rule 
we obtained 4.85 inches). 

309. Radii of Gyration. — The following table, taken from 
p. 523 of Rankine's Civil Engineering, gives values of Jc 2 , 
the square of the least radius of gyration of the given cross- 
section about a gravity-axis. By giving the least value of 



FLEXUKE. LONG COLUMNS. 



377 



¥ it is implied that the plane of flexure is not determined 
by the end-conditions of the column ; (i.e., it is implied 
that the column has either flat ends or round ends.) If 
either end (or both) is a pin-joint the column may need to 
be treated as having a flat-end as regards flexure in a plane 
containing the axis of the column and the axis of the pin, 
if the bearings of the pin are firm ; while as regards flex- 
ure in a plane perpendicular to the pin it is to be consid- 
ered round -ended at that extremity. 

In the case of a " thin cell " the value of ft? is strictly 
true for metal infinitely thin and of uniform thickness ; still, 
if that thickness does not exceed yi of the exterior diame- 
ter, the form given is sufficiently near for practical pur- 
poses ; similar statements apply to the branching forms. 



(a) 



ii 



(5) 



* 5- 




Fig. 317. 





Solid Bectangle. 
h= least side. 
Thin Square Cell. 
Side= k 



Fig. 317(a). V=±W 
Fig. 317 (b\ k 2 =U 2 



Thin Kect angular Cell. -p. q-i 7 



h= least side. 

Solid Circular Section. 

Diameter —d. 

Thin Circular Cell. 

Exterior diam, = d. 



»• 



h 2 h+3b 



12 h+b 

Fig. 317(d). V=htf 

Fig. 317 (e). &=Ld* 



Angle-Iron of Equal -pie. S17 ( f) ¥=±.& 



ribs 



378 



MECHANICS OF ENGINEERING. 



Angle-Iron of unequal 

ribs. 
Cross of equal arms. 
I-Beam as a pillar. 
Let area of web =B. 

" " both flanges 

=A. 
Channel 
Iron. 



¥h 2 



Fig. 318 (a). V= ^ 
Tig. 318 (6), W=\b 2 



Fig. 318 (c). k 2 = 



A 



12 A+B 



Fig. 318 (c/). ¥=h 2 



■+- 



^?_1 



L12 (^+J?) 4 (^+i?) 2 J 
Let area of web =B ; of flanges =A (both), h extends 
from edge of flange to middle of web. 





Fig. 319. 



PHCENIX COLUMN. 



Fig. 330. 



310. Built Columns.— The " compression members " of 
wrought-iron bridge trusses are generally composed of 
several pieces riveted together, the most common forms 
being the Phoenix column (ring-shaped, in segments,) and 
combinations of channels, plates, and lattice, some of which 
are shown in Figs. 319 and 320. 

Experiments on full size columns of these kinds were 
made by the U. S. Testing Board at the Watertown Arse- 
nal about 1880. 

The Phoenix columns ranged from 8 in. to 28 feet in 
length, and from 1 to 42 in the value of the ratio of length 
to diameter. The breaking loads were found to be some- 
what in excess of the values computed from Bankine's 
formula ; from 10 to 40 per cent, excess. In the pocket- 
book issued by the Phoenix company they give the follow- 
ing formula for their columns, (wrought-iron.) 



FLEXUKE. LONG COLUMNS. 379 

Breaking load in lbs. ) 50,000 F m 

for flat-ended columns )~r+ V W 

3,000ft 2 

where F = area in sq. in., I = length, and h = external 
diameter, both in the same unit. 

Many different formulae have been proposed by different 
engineers to satisfy these and other recent experiments on 
columns, but all are of the general form of Rankine's* 
For instance Mr. Bouscaren, of the Keystone Bridge Co., 
claims that the strength of Phoenix columns is best given 
by the formula 

Breaking load in ) = 38,000 F ,o\ 

lbs. for flat-ends, j tf V< 

1 + 100,000& a 

(F must be in square inches.) 

The moments of inertia, I, and thence the value of k 2 —■ 
I -7- F, for such sections as those given in Figs. 319 and 
320 may be found by the rules of §§ 85-93, (see also § 258.) 

311. Moment of Inertia of Built Column. Example. — It is pro- 
posed to form a column by joining two I-beams by lattice- 
work, Fig. 321, (a). (While the lattice-work is relied upon 
to cause the beams to act together as one piece, it is not 
regarded in estimating the area F, or the moment of iner- 
tia, of the cross section). It is also required to find the 
proper distance apart = x, Fig. 321, at which these beams 
must be placed, from centre to centre of webs, that the 
liability to flexure shall be equal in all axial planes, i.e. 
that the 1 of the compound section shall be the same 
about all gravity axes. This condition will be ful- 
filled if I Y can be made = I x * (§89), being the centre 
of gravity of the compound section, and X perpendicular 
to the parallel webs of the two equal I-beams. 

Let F r = the sectional area of one of the I-beams, F x 
(see Fig. 321(a) its moment of inertia about its web-axis, i" x ' 
that about an axis "] to web. (These quantities can be 

* That is, with flat ends or ball ends ; but with pin ends, Fig. 312, if the 
pin is || to X, put 47y = ix ; if II to T, put 4Jx = Iy . 



380 



MECHANICS OF ENGINEERING. 



found in the hand-book of the iron company, for each size 
of rolled beam). 
Then the 

total I x = 2I' X ; and total 7 y = 2[> v + W-Y] 

(see §88 eq. 4.) If these are to be equal, we write them so 
and solve for x, obtaining 



/ 4[JWv 

v — J* — 



(1) 



312. Numerically; suppose" each girder to be a 10 j4 inch 
light I-beam, 105 lbs. per yard, of the N. J. Steel and Iron 
Co., in whose hand-book we find that for this beam Z' x = 
185.6 biquad. inches, and I\ = 9.43 biquad. inches, while 
F' = 10.44 sq. inches. With these values in eq. (1) we 
have 



U (185.6-9.43) _ VW75 = a21 inches> 

V 10 4-4 



!v 

SIT 7 



-X—4 



(a) 




Fiq. 321. 



The square of the radius of gyration will be 
k 2 =2F x +2F'= 371.2 -^20.88=17.7 sq. in. 



(2) 



and is the same for any gravity axis (see § 89). 

As an additional example, suppose the two I-beams united 
by plates instead of lattice. Let the thickness of the plate 
= t, Fig. 321, (b). Neglect the rivet-holes. The distance 
a is known from the hand-book. The student may derive 
a formula for x, imposing the condition that (total ix) = ^ 



FLEXURE. LONG COLUMNS. 



381 



313. Trussed Girders. — When a horizontal beam is trussed 

W=-2 Wl 







Fig. 323. 

in the manner indicated in Fig. 322, with a single post or 
strut under the middle and two tie-rods, it is subjected to 
a longitudinal compression due to the tension of the tie- 
rods, and hence to a certain extent resists as a column, the 
plane of whose flexure is vertical, (since we shall here sup- 
pose the beam supported laterally .)Taking the case of uni- 
form loading, (total load = W)snid supposing the tie-rods 
screwed up (by sleeve nuts) until the top of the post is on 
a level with the piers, we know that the pressure between 
the post and the beam is P' = $/§ W (see § 273). Hence 
by the parallelogram of forces (see Fig. 322) the tension 
in each tie-rod is 



2cos« 16 " cos a 

At each pier the horizontal component of Q is 

5 
P= Q sin a—— - Wt&na 
16 



(i) 



Hence we are to consider the half -beam 50asa" pin-and- 
square " column under a compressive force P= 5 /m W tan a f 
as well as a portion of a continuous girder over three 
equidistant supports at the same level and bearing a uni- 
form load TV. In the outer fibre of the dangerous section, 
0, (see also § 273 and Fig. 278) the compression per sq. 
inch due to both these straining actions must not exceed 
a safe limit, B', (see § 251). In eq. (6) § 307, where P 2 is 
tne breaking force for a pin-and-square column, the great- 



382 MECHANICS OF ENGINEERING. 

est stress in any outer fibre = G ( == the Modulus of Crush* 
ing) per unit of area. If then we write p Q(A . instead of G 
in that equation, and 5 / 16 Wtsm. a instead of P 2 we have 

max. stress due ) __ 5 Wt&nar.. . 16 o I 2 ~|. 

to column action j ~ P™\— jg • -p 1 ~r -g-* r '"p J> 

while from eq. (3), p. 326, we have (remembering that our 
present W represents double the W oi §273). 

max. stress due J = = 1 Wle^ 1 Wle 
to girder action \ ^ G[ Jg J " Jg * ^pp 

By writing p co i.-\-p S i= R'= a safe value of compression per 
unit-area, we have the equation for safe loading 

W[5t.na(l+™. ■{>**}+ kyUFlV . . (2) 

Here I = the half-span OB, Fig. 322, e = the distance of 
outer fibre from the horizontal gravity axis of the cross 
section, k 2 = the radius of gyration of the section referred 
to the same axis, while F = area of section. /9 should be 
taken from the end of §307. 

Example.— If the span is 30 ft. = 360 in., the girder a 15 
inch heavy I-beam of wrought iron, 200 lbs. to the yard, in 
which e = T / 2 of 15 = iy 2 inches, .F=20 sq. in., and k 2 = 
35.3 sq. inches (taken from the Trenton Co.'s hand-book), 
required the safe load W, the strut being 5 ft. long. 
From §307, ft = 1 : 36,000 ; tan a = 15--5 = 3.00. Hence, 
using the units pound and inch throughout, and putting 
R' = 12,000 lbs. per sq. in. = max. allowable compression 
stress, we have from eq. (2) 

w _ 16x20x12,000 =71 1 - 1 ,, 

15fl + 16 1 m im 2 l, lS0x7y 2 ='35.5 tons. 
L 1 " 1- 9 36,000 ' 35.3 J" 1 " 35.3 

i. e., 69,111 lbs. besides the weight of the beam. 

If the middle support had been a solid pier, the safe load 
would have been 48 tons ; while if there had been no 
middle support of any kind, the beam would bear safelv 



FLEXURE. LOXG COLUMXS. 



383 



only 11.5 tons, 
the strut)]. 



[Let the student design the tie-rods (and 



314. Buckling of Web-Plates in Built Girders. — In §257 men- 
tion was made of the fact that very high web plates in 
built beams, such as / beams and box-girders, might need 
to be stiffened by riveting T-irons on the sides of the web. 
(The girders here spoken of are horizontal ones, such as 
might be used for carrying a railroad over a short span of 
20 to 30 feet. 

An approximate method of determining whether such 
stiffening is needed to prevent lateral buckling of the web, 
may be based upon Rankine's formula for a long column 
-and will noAv be given. 

In Fig. 323 we have, free, a portion of a bent I-beam, 
between two vertical sections at a distance apart= h Y = 
the height of the web. In such a beam under forces |_ to 
its axis it has been proved (§256) that we may consider 
the web to sustain all the shear, J, at any section, and the 
flanges to take all the tension and compression, which 
form the " stress -couple" of the section. These couples 
and the two shears are shown in Fig. 323, for the two 
exposed sections. There is supposed to be no load on this 
portion of the beam, hence the shears at the two ends are 




FlU. 323. 



Fig. 325. 



equal. Now the shear acting between each flange and the 
horizontal edge of the web is equal in intensity per square 
inch to that in the vertical edge of the web ; hence if the 
web alone, of Fig. 323, is shown as a free body in Fig. 324, 
we must insert two horizontal forces = J, in opposite 



384 MECHANICS OF ENGINEERING. 

directions, on its upper and lower edges. Each of these 
= J since we have taken a horizontal length \ = height 
of web. In this figure, 324, we notice that the effect of 
the acting forces is to lengthen the diagonal BD and 
shorten the diagonal AC, both of those diagonals making 
an angle of 45° with the horizontal. 

Let us now consider this buckling tendency along AC, 
by treating as free the strip AC, of small width = b v This 
is shown in Fig. 325. The only forces acting in the direc- 
tion of its length AC are the components along AC of the 
four forces J' at the extremities. We may therefore treat 
the strip as a long column of a length I = h t a/2, of a sec- 
tional area F = bb l9 (where b is the thickness of the web 
plate), with a value of h 2 = 1 / 12 b 2 (see § 309), and with 
fixed (or flat) ends. Now the sum of the longitudinal 
components of the two JVs at A is Q — 2 J' y 2 V2 

= J' a/2 ; but J' itself — tt. b ]/ 2 b x a/2, since the small 

rectangle on which J' acts has an area = b y 2 b x a/2, and 
the shearing stress on it has an intensity of (J -*- bhq) per 
unit of area. Hence the longitudinal force at each end of 
this long column is 

Q = h if » 

According to eq. (4) and the table in § 307, the safe load 
(factor of safety = 4) for a wrougM-iron column of this 
form, with flat ends, would be (pound and inch) 

p = #&6 1 36,000 9,000 bb x m 

1 i+i w i . l v ' • ■ * K) 

36,000 Vi# 1,500 6 2 . 

If, then, in any particular locality of the girder (of 
wrought-iron) we find that Q is > P lt i.e. 

if ^ is > 9 ' 0005 ■ (pound and inch) . , (3). 
fii I hi 

1+ T~oW!? 






FLEXUBE. LONG COLUMNS. 



%85 



then vertical stiffeners will be required laterally. 

When these are required, they are generally placed at 
intervals equal to h lt (the depth of web), along that part 
of the girder where Q is > P x . 

Example Fig. 326. — Will stiffening pieces be required 
in a built girder of 20 feet span, bearing a uniform load of 
40 tons, and having a web 24 in. deep and j/fr in. thick ? 

From § 242 we know that the 
greatest shear, J max., is close to 



W =40 TONS 



either pier, and hence we investi- 1 k — -io-— -^ 
gate that part of the girder first. 

J max. = y 2 W = 20 tons 
=40,000 lbs. 
,*, (inch and lb.), see (3), 







J 40,000 



Fig 326. 



h 



24 



1666.6 (4) 



while, see (3), (inch and pound), 



9,000 x^ 



1+ 



24 2 



=905.0 



(5) 



1,500 



which is less than 1666.66. 

Hence stiffening pieces will be needed near the extremi- 
ties of the girder. Also, since the shear for this case of 
loading diminishes uniformly toward zero at the middle 
they will be needed from each end up to a distance of 
ft of 10 ft. from the middle. 



386 MECHANICS OF EXGItfEERIXG. 



CHAPTEE VII. 



lilNEAR ARCHES (OF BLOCKWORK), 



315. A Blockwork Arch is a structure, spanning an opening 
or gap, depending, for stability, upon the resistance to 
compresssion of its blocks, or voussoirs, the material of 
which, such as stone or brick, is not suitable for sustain- 
ing a tensile strain. Above the voussoirs is usually 
placed a load of some character, (e.q. a roadway,) whose 
pressure upon the voussoirs will be considered as vertical, 
only. This condition is not fully realized in practice, 
unless the load is of cut stone, with vertical and horizontal 
joints resting upon voussoirs of corresponding shape (see 
Fig. 327), but sufficiently so to warrant 
its assumption in theory. Symmetry 
of form about a vertical axis will also 
be assumed in the following treatment. 




316. Linear Arches. — For purposes of 
theoretical discussion the voussoirs of 
Fig. 327 may be considered to become 
fig. 327. infinitely small and infinite in number, 

thus forming a " linear arch," while retaining the same 
shapes, their depth "1 to the face being assumed constant 
that it may not appear in the formulae. The joints 
between them are "| to the curve of the arch, i.e., adjacent 
voussoirs can exert pressure on each other only in the 
direction of the tangent-line to that curve. 



LIKEAK AECHES. 



387 



317. Inverted Catenary, or Linear Arch Sustaining its Own 
Weight Alone. — Suppose the infinitely small voussoirs to 
have weight, uniformly distributed along the curve, weigh- 
ing q lbs. per running linear unit. The equilibrium of 
such a structure, Fig. 328, is of course unstable but theo- 
retically possible. Required the form of the curve when 
equilibrium exists. The conditions of equilibrium are, 
obviously : 1st. The thrust or mutual pressure T between 
any two adjacent voussoirs at any point, A, of the curve 
must be tangent to the curve ; and 2ndly, considering a 
portion BA as a free body, the resultant of H the pres- 




Fig. 328. 



Fig. 329. 



Fig. 330. 



sure at B the crown, and T at A, must balance R the re- 
sultant of the || vertical forces (i.e.,weights of the elementary 
voussoirs) acting between B and A. 

But the conditions of equilibrium of a flexible, inexten- 
sible and uniformly loaded cord or chain are the very 
same (weights uniform along the curve) the forces being 
reversed in. direction. Fig. 329. Instead of compression 
we have tension, while the || vertical forces act toward in- 
stead of away from, the axis X. Hence the curve of equi- 
librium of Fig. 328 is an inverted catenary (see § 48) whose 
equation is 



y+c=y 2 c^ e c + e ..] 



(1) 



See Fig. 330. e = 2.71828 the Naperian Base. The "par- 
ameter" c may be determined by putting, x = a, the half 
span, and y= Y, the rise, then solving for c by successive 



388 



MECHANICS OF ENGINEERING. 



approximations. The " horizontal thrust" or H , is — qc, 
while if s = length of arch OA, along the curve, the thrust 
T at any point A is 



T=VH 2 +q 2 s 2 (2.) 

From the foregoing it may be inferred that a series of vous- 

soirs of finite dimensions, arranged 

so as to contain the catenary curve, 

with joints "| to that curve and of 

equal weights for equal lengths of 

arc will be in equilibrium, and 

moreover in stable equilibrium on 

account of friction, and the finite 

width of the joints ; see Fig. 331, 




FIG. 331. 



318. Linear Arches under Given Loading. — The linear arches 
to be considered further will be treated as without weight 
themselves but as bearing vertically pressing loads (each 
voussoir its own). 

Problem. — Given the form of the linear arch, itself, it is 
required to find the law of vertical depth of loading under 
which the given linear arch will be in equilibrium. Fig. 
332, given the curve ABC, i.e., the linear arch itself, re- 
quired the form of the curve MON, or upper limit of load- 
ing, such that the linear arch ABC shall be in equilibrium 
under the loads lying between the two curves. The load- 
ing is supposed homogeneous and of constant depth ~| to 
paper ; so that the ordinates z between the two curves are 
proportional to the load per horizontal linear unit. Assume 
a height of load z at the crown, at pleasure ; then required 
the z of any point m as a function of z and the curve 
ABC. 




Fig. 332 



LIXEAR ARCHES. 



389 



Practical Solution. — Since a linear arch under vertical 
pressures is nothing more than the inversion of the curve 
assumed by a cord loaded in the same way, this problem 
might be solved mechanically by experimenting with a 
light cord, Fig. 333, to which are hung other heavy cords, 
or bars of uniform weight per unit length, and at equal 
horizontal distances apart ivhen in equilibrium. By varying 
the lengths of the bars, and their points of attachment, we 
may finally find the curve sought, MON. (See also § 343.) 

Analytical Solution. — Consider the structure in Fig. 334 
A number of rods of finite length, in the same plane, are in 
equilibrium, bearing the weights P, P l9 etc., at the con- 




Fig. 334. 



Fig. 335. 



necting joints, each piece exerting a thrust T against the 
adjacent joint. The joint A 9 (the " pin " of the hinge), im- 
agined separated from the contiguous rods and hence free, 
is held in equilibrium by the vertical force P (a load) and 
the two thrusts T and T' 9 making angles = d and d' with 
the vertical ; Fig. 335 shows the joint A free. From ^(hor^ 
izontal comps.)=0, we have. 

rsin#=T'sin0'. 

That is, the horizontal component of the thrust in any rod 
is the same for all ; call it H Q . .*. 



T= 



sin 



a) 



390 



MECHANICS OF ENGINEERING. 



Now draw a line As T to T p and write 2 ( compons. || to 
As)=0 ; whence P sin d'= T sin fi, and [see (1)] 



p _ H sin p 
sin sin 0' 



(2) 



Let the rods of Fig. 334 become infinitely small and infi- 
nite in number and the load continuous. The length of 
each rod becomes =ds an element of the linear arch, ft is 
the angle between two consecutive efo's, is the angle be- 
tween the tangent line and the vertical, while P becomes 
the load resting on a single dx, or horizontal distance be- 
tween the middles of the two cfo's. That is, Fig. 336, if 
y= weight of a cubic unit of the 
loading, P—yzdx. (The lamina of 
arch and load considered is unity, 
T to paper, in thickness.) H =& 
constant = thrust at crown ; 
0=0', and sin fi=dss-p, (since the 
angle between two consecutive tan- 
gents is = that between two con- 
secutive radii of curvature). Hence 
eq. (2) becomes 



yzdx 



H n ds 



p sin 2 



but dx—ds sin 0, 




Fig. 336. 



yz: 



H n 



p sin 3 



(3) 



Call the radius of curvature at the crown p 0> and since 
there z= z and 0=90°, (3) gives pbPo^Hoi hence (3) may 
be written 



_ V> 



sin 3 # 



W 



This is the law of vertical depth of loading required. For 
a point of the linear arch where the tangent line is verti- 
cal, sin #=0 and z would = oo ; i.e., the load would be in- 



LINEAR ARCHES. 



391 



finitely high. Hence, in practice, a full semi-circle, for in- 
stance, could not be used as a linear arch. 

319, Circular Arc as Linear Arch. — As an example of the 

preceding problem let us ap- 
ply eq. (4) to a circular arc, 
Fig. 337, as a linear arch. 
Since for a circle p is con- 
stant — r, eq. (4) reduces 
to 




z~ 



HS? • (5) 



Hence the depth of loading 
must vary inversely as the cube of the sine of the angle 6 
made by the tangent line (of the linear arch) with the ver- 
tical. 

To find the depth z by construction.— Having z given, 
being the centre of the arch, prolong Ga and make ab = 
z ; at b draw a "] to Gb, intersecting the vertical through a 
at some point d ; draw the horizontal dc to meet Ga at 
some point c. Again, draw ce ~| to (7c, meeting ad in e ; 
then ae = z required ; a being any point of the linear arch. 
For, from the similar right triangles involved, we have 

z =ab=ad sin d—ac sin 6. sin d=ae sin d sin 6 sin 



ae 



sin 3 # ' 



i.e., ae- 



■z. Q.E.D. 

[see (5.)] 



320. Parabola as Linear Arch. — To apply eq. 4 § 318 to a 

parabola (axis vertical) as linear arch, we must find values 
of f) and p the radii of curvature at any point and the 
crown respectively, That is, in the general formula, 



■■Mm 



dx 



we must substitute the forms for the first and second dif- 
ferential co-efficients, derived from the equation of the 



392 



MECHANICS OF ENGINEERING. 





Fig. 338. 



Fig. 339. 



curve (parabola) in Fig. 338, i.e. from x 2 = 2 py ; whence 
we obtain 

-^,or cot #,= _and^4-=— 
ax p axr p 



Hence p a Av ~\ 1= p cosec. 3 #, i.e. p 

1-7-p 



= P 

sin 3 # 



(6) 



At the vertex d == 90° .*. p = p. Hence by substituting 
for p and p in eq. (4) of § 318 we obtain 

z=z = constant [Fig. 339 J (7) 

for a parabolic linear arch. Therefore the depth of homo- 
geneous loading must be the same at all points as at the 
crown ; i.e., the load is uniformly distributed with respect 
to the horizontal. This result might have been antici- 
pated from the fact that a cord assumes the parabolic 
form when its load (as approximately true for suspension 
bridges) is uniformly distributed horizontally. See § 46 
in Statics and Dynamics. 

321. Linear Arch for a Given Upper Contour of Loading, the 
arch itself being the unknown lower contour. Given the 
upper curve or limit of load and the depth z at crown, re- 
quired the form of linear arch which will be in equili- 
brium under the homogenous load between itself and that 
upper curve. In Fig. 340 let MON be the given upper 
contour of load, z is given or assumed,*;' and z" are the 
respective ordinates of the two curves BA G and MON 
Eequired the eqation of BAG 






LINEAR AECHES. 



393 




Fig. 340. 



Fig 341. 



As before, the loading is homogenous, so that the 
weights of any portions of it are proportional to the 
corresponding areas between the curves. (Unity thick- 
ness T to paper.) Now, Fig. 341, regard two consecutive 
ds's of the linear arch as two links or consecutive blocks 
bearing at their junction m the load dP = y (z + z"} dx in 
which y denotes the heaviness of weight of a cubic unit of 
the loading. If T and T are the thrusts exerted on these 
two blocks by their neighbors (here supposed removed) 
we have the three forces dP, T and T' } forming a system 
in equilibrium. Hence from IX =0. 



T cos (p — T cos qf 



iind 



IY=0 gives T sin cp' — T sin <p = dP 



(1) 



(2) 



From (1) it appears that T cos <p is constant at all points 
of the linear arch (just as we found in § 318) and hence 
= the thrust at the crown, = H, whence we may write 

T=H + cos tp and T=H + cos <p' . . . (3) 

Substituting from (3) in (2; we obtain 

H (tan <p' — tan (p)—dP (4) 

But tan <p f = — - and tan <p' — — ~X- , (dx constant) 

dx dx 

while dP = y (z f + O dx. Hence, putting for convenience 

H. — ya 2 , (where a — side of an imaginary square of the 



394 MECHANICS OF ENGINEERING. 

loading, whose thickness = unity and whose weight = H) 
we have. 

£=>'+*"> ( 5 > 

as a relation holding good for any point of the linear arch 
which is to be in equilibrium under the load included 
between itself and the given curve whose ordinates are z", 
Fig. 340. 

322. Example of Preceding. Upper Contour a Straight Line.— 
Fig. 342. Let the upper contour be a right line and hor- 
izontal ; then the z" of eq. 5 becomes zero at all points of 
ON. Hence drop the accent of z' in eq. (5) and we have 

d 2 z = z 
dx 2 a 2 

Multiplying which by dz we obtain 

dz dh 1 



da? a 2 



zdz (6) 



This being true of the z, dz, d 2 z and dx of each element of 
the curve O'B whose equation is desired, conceive it writ- 
ten out for each element between 0' and any point m, and 
put the sum of the left-hand members of these equations 
= to that of the right-hand members, remembering that 
a 2 and dx 2 are the same for each element. This gives 

dz—dz z^z 

' dx 2 / a 2 I ' 'da? 2 a 2 |_2 2 J 

d(JL) 

agg _ n V *q/ . . . . (7.) 



dx= =«. 



Vz 2 -z 2 I 



'© 



LLNEAB ABCHES. 



395 




Fig. 342. Fig : 343 

Integrating (7.) between 0' and any point m 



[■ 



/, 



-l^m- 1 ) 



i.e., x=a log. 



[ 



Md' 



] 



or z 



= f [«W] 



(8)' 



(8.) 



(9.) 



This curve is called the transformed catenary since we may 
obtain it from a common catenary by altering all the orcli- 
nates of the latter in a constant ratio, just as an ellipse 
may be obtained from a circle. If in eq. (9) a were = z Q 
the curve would be a common catenary. 

Supposing z and the co-ordinates x x and z l of the point 
B (abutment) given, we may compute a from eq. 8 by put- 
ting x =x l and z = z lf and solving for a. Then the crown- 
thrust H = ya 2 becomes known, and a can be used in eqs. 
(8) or (9) to plot points in the curve or linear arch. From 
eq. (9) we have 

(10) 



area 
00' mn 



\ =J zdx=2j [e dx+e cfoJ=TL e "" e J 



Fig. 343. 

Call this area, A. As for the thrusts at the different 
joints of the linear arch, see Fig. 343, we have crown- 
thrust =H = ya 2 . . ; (11) 
and at any joint m the thrust 

T=^/H 2 +{ r A) 2 =rVtf+A 2 .... (12) 



396 MECHANICS OF ENGINEERING. 

323. Remarks. — The foregoing results may be utilized 
with arches of finite dimensions by making the arch-ring 
contain the imaginary linear arch, and the joints "[ to the 
curve of the same. Questions of friction and the resist- 
ance of the material of the voussoirs are reserved for a 
succeeding chapter, (§ 344) in which will be advanced a 
more practical theory dealing with approximate linear 
arches or " equilibrium polygons " as they will then be 
called. Still, a study of exact linear arches is valuable on 
many accounts. By inverting the linear arches so far pre- 
sented we have the forms assumed by flexible and inexten- 
sible cords loaded in the same way. 



GBAPHICAL STATICS. 397 



CHAPTEE YnL 



ELEMENTS OF GRAPHICAL, STATICS. 



324. Definition. — In many respects graphical processes 
have advantages over the purely analytical, which recom- 
mend their use in many problems where celerity is desired 
without refined accuracy. One of these advantages is that 
gross errors are more easily detected, and another that 
the relations of the forces, distances, etc., are made so 
apparent to the eye, in the drawing, that the general effect 
of a given change in the data can readily be predicted at 
a glance. 

Graphical Statics ia the system of geometrical construc- 
tions by which problems in Statics may be solved by 
the use of drafting instruments, forces as well as distances 
being represented in amount and direction by lines on the 
paper, of proper length and position, according to arbi- 
trary scales ; so many feet of distance to the linear inch of 
paper, for example, for distances ; and so many pounds or 
tons to the linear inch of paper for forces. 

Of course results should be interpreted by the same 
scale as that used for the data. The parallelogram of 
forces is the basis of all constructions for combining and 
resolving forces. 

325. Force Polygons and Concurrent Forces in a Plane. — If a 
material point is in equilibrium under three forces P x P 2 
P 3 (in the same plane of course) Fig. 344, any one of them, 



398 



MECHANICS OF ENGINEERING. 




as P lf must be equal and opposite to B the resultant of 
the other two (diagonal of their parallelogram). If now 
we lay off to some convenient scale a line in Fig. 345 = 
P x and II to P x in Fig. 344 ; and then from the pointed end 

of Pi a line equal and || to P 2 and 
laid off pointing the same ivay, we 
note that the line remaining to 
P close the triangle in Fig. 345 must 
be = and || to P 3 , since that tri- 
angle is nothing more than the 
left-hand half-parallelogram of 
Fig. 345. Fig. 344. Also, in 345, to close 
the triangle properly the directions of the arrows must 
be continuous Point to Butt, round the periphery. Fig. 
345 is called a force polygor ; ; of three sides only in this 
case. By means of it, given any two of the three forces 
which hold the point in equilibrium, the third can be 
found, being equal and || to the side necessary to " close " 
the force polygon. 

Similarly, if a number of forces in a plane hold a mate- 
rial point in equilibrium, Fig. 346, their force polygon, 




Fig.344. 





Fig. 347, must close, whatever be the order in which its 
sides are drawn. For, if we combine P x and P 2 into a re- 
sultant 0a 9 Fig. 346, then this resultant with P 3 to form a 
resultant Ob, and so on ; we find the resultant of P l9 P 2 , P 3 , 
and P 4 to be Oc, and if a fifth force is to produce equilib- 
rium it must be equal and opposite to Oc, and would close 
the polygon OdabcO, in which the sides are equal and par- 



GKAPHICAL STATICS. 



399 



allel respectively to the forces mentioned. To utilize this 
fact we can dispense with all parts of the parallelograms in 
Fig. 346 except the sides mentioned, and then proceed as 
follows in Fig. 347 : 

If P 5 is the unknown force which is to balance the other 
four (i.e, is their anti -resultant), we draw the sides of the 
force polygon from A round to B, making each line paral- 
lel and equal to the proper force and pointing the same 
way ; then the line BA represents the required P 5 in 
amount and direction, since the arrow BA must follow 
ihe continuity of the others (point to butt). 

If the arrow BA were pointed at the extremity B, then 
it gives, obviously, the amount and direction of the result- 
ant of the four forces P 1 . . . P 4 . The foregoing shows 
that if a system of Concurrent Forces in a Plane is in equi- 
librium, its force polygon must close. 



326. Non-Concurrent Forces in a Plane, — Given a system of 
non-concurrent forces m a plane, acting on a rigid body, 
required graphic means of finding their resultant and anti- 
resultant ; also of expressing conditions of equilibrium. 
The resultant must be found in amount and direction ; and 
also in position (i.e., its line of action must be determined). 
E. g., Fig. 348 shows a curved rigid beam fixed in a vise 
at P, and also under the action of forces P x P 2 P 3 and P 4 
(besides the action of the vise); required the resultant of 

P 17 P 2 , P 3 , and P 4 . 
By the ordinary 
parallelogram of 
forces we com- 
bine P x and P 2 at 
a, the intersection 
of their lines of 
action, into a re- 
sultant P a ; then P a with P 3 at b, to form P b ; and finally P b 
with P 4 at c to form R c which is .*. the resultant required, 
i.e., of P x . . . . P 4 ; and c . . . F is its line of action. 




<±vO 



MECHANICS OF ENGINEERING. 




Fig. 349. 



The separate force triangles (half-parallelograms) by 
which the successive partial resultants P a , etc., were found, 
are again drawn in Fig. 349. Now since P c , acting in the 

line c.F, Fig. 348, 
is the resultant of 
f Pi., P 4 , it is plain 
* that a force 22/ 
equal to P c and act- 
ing along c. . P,but 
in the opposite di- 
rection, would balance the system P l . . . P 4 , (is their anti- 
resultant). That is, the forces P x P 2 P 3 P 4 and P/ would 
form a system in equilibrium. The force 2?/ then, repre- 
sents the action of the vise T upon the beam. Hence re- 
place the vise by the force P c ' acting in the line . . . F . . . c j 
to do which requires us to imagine a rigid prolongation of 
that end of the beam, to intersect F . . . c. This is shown in 
Fig. 350 where the whole beam is free, in equilibrium, under 
the forces shown, and in precisely the same state of stress, 
part for part, as in Fig. 348. Also, by combining in one 
force diagram, in Fig. 351, all the force triangles of Fig. 349 
(by making their common sides coincide, and putting P c ' 
instead of P c , and dotting all forces other than those of 
Fig. 350), we have a figure to be interpreted in connection 
with Fig, 350. 




SPACE DIAGRAM 
Fig. 350. 



FORCE DIAGRAM 
Fig. 351. 



Here we note, first, that in the figure called a force-dia- 
gram, P 1 P 2 P 3 P 4 and 22/ form a closed polygon and that 



GRAPHICAL STATICS. 40 1 

their arrows follow a continuous order, point to butt, 
around the perimeter ; which proves that one condition of 
equilibrium of a system of non-concurrent forces in a plane 
is that its force polygon must close. Secondly, note that ab 
is || to Oa' , and be to Ob' ; hence if the force-diagram has 
been drawn (including the rays, dotted) in order to deter- 
mine the amount and direction of P c ', or any other one force, 
we may then find its line of action in the space-diagram, as 
follows: (N. B. — By space diagram is meant the figure show- 
ing to a true scale the form of the rigid body and the lines 
of action of the forces concerned). Through a, the intersec- 
tion of P 1 and P 2 , draw a line || to Oa' to cut P 3 in some point 
b ; then through b a line || to Ob' to cut P 4 at some point c; cF 
drawn || to Oc' is the required line of action of P c ', the anti- 
resultant of P x , P 2 , P 3 , and P 4 . 

abc is called an equilibrium polygon; this one having but 
two segments, ab and be (sometimes the lines of action of P l 
and P c ' may conveniently be considered as segments.) The 
segments of the equilibrium polygon are parallel to the respect- 
ive rays of the force diagram. 

Hence for the equilibrium of a system of non-concurrent 
forces in a plane not only must its force polygon close, 
but also the first and last segments of the corre- 
sponding equilibrium polygon must coincide with 
the resultants of the first two forces, and of the last 
two forces, respectively, of the system. E.g., ab coin- 
cides with the line of action of the resultant of P x and P 2 ; 
he with that of P 4 and B/ c . Evidently the equil. polygon 
will be different with each different order of forces in 
the force polygon or different choice of a pole, 0. But if 
the order of forces be taken as above, as they occur along 
the beam, or structure, and the pole taken at the " butt " of 
the first force in the force polygon, there will be only one ; 
(and this one will be called the special equilibrium polygon 
in the chapter on arch-ribs, and the " true linear arch " in 
dealing with the stone arch.) After the rays (dotted in 
Fig. 351) have been added, by joining the pole to each 



402 MECHANICS OF ENQENEERXXG. 

vertex with which it is not already connected, the final 
figure may be called the force diagram. 

It may sometimes be convenient to give the name of 
rays to the two forces of the force polygon which meet 
at the pole, in which case the first and last segments of 
the corresponding equil. polygon will coincide with the 
lines of action of those forces in the space-diagram (as we 
may call the representation of the body or structure on 
which the forces act). This " space diagram " shows the 
real field of action of the forces, while the force diagram, 
which may be placed in any convenient position on the 
paper, shows the magnitudes and directions of the forces 
acting in the former diagram, its lines being interpreted 
on a scale of so many lbs. or tons to the inch of paper ; in 
the space-diagram we deal with a scale of so m&nyfeet to 
the inch of paper. 

We have found, then, that if any vertex or corner of the 
closed force polygon be taken as a pole, and rays drawn 
from it to all the other corners of the polygon, and a cor- 
responding equil. polygon drawn in the space diagram, the 
first and last segments of the latter polygon must co-incide 
with the first and last forces according to the order 
adopted (or with the resultants of the first two and last 
two, if more convenient to classify them thus). It remains 
to utilize this principle. 

327. To Find the Resultant of Several Forces in a Plane. — This 

might be done as in § 326, but since frequently a given set 

of forces are parallel, or nearly so, a special method will 

now be given, of great convenience in such cases. Fig. 352. 

\ d Let Pi P 2 and 

/ 1 ,\ /! >si forces whose 

/ \ \ p/ l "M 

. v w / \f' \ resultant is re- 

/£\ // \ y quired. Let us 

\ P ' \^4^ \ first find their 

\ p\ 1 / 2 anCi - resultant, 

^£ ?3 or force which 

Fig. 352. Pw. 353. will balance 



GRAPHICAL STATICS. 403 

them. This anti-resultant may be conceived as decom- 
posed into two components P and P" one of which, say P, 
is arbitrary in amount and position. Assuming P, then, 
at convenience, in the space diagram, it is required to find 
P'. The five forces must form a balanced system ; hence 
if beginning at lt Fig. 353, we lay off a line O x A = P by 
scale, then A\ = and || to P { , and so on (point to butt), the 
line BO Y necessary to close the force polygon is = P' re- 
quired. Now form the corresponding equil. polygon in 
the space diagram in the usual way, viz.: through a the 
intersection of P and P l draw ab || to the ray O i ... 1 
(which connects the pole O l with the point of the last force 
mentioned). From b, where ab intersects the line of P 2 , 
draw be, II to the ray 1 . . 2, till it intersects the line of P 3 . 
A line me drawn through c and || to the P' of the force 
diagram is the line of action of P'. 

Now the resultant of P and P' is the anti-resultant of 
P 1? P 2 and P 3 ; .*. d, the intersection of the lines of P and 
P', is a point in the line of action of the anti-resultant re- 
quired, while its direction and magnitude are given by the 
line BA in the force diagram ; for BA forms a closed poly- 
gon both with P L P 2 P 3 , and with PP\ Hence a line 
through d || to BA, viz., de, is the line of action of the anti- 
resultant (and hence of the resultant) of P lf P 2 , P 3 . 

Since, in this construction, P is arbitrary, we may first 
choose Oi, arbitrarily, in a convenient position, i.e., in such 
a position that by inspection the segments of the result- 
ing equil. polygon shall give fair intersections and not 
pass off the paper. If the given forces are parallel the 
device of introducing the oblique P and P' is quite neces- 
sary. 

328. — The result of this construction may be stated as 
follows, (regarding Oa and cm as segments of the equil. 
polygon as well as ab and be): If any two segments of an 
equil. polygon be prolonged, their intersection is a point in 
the line of action of the resultant of those forces acting at 



404 



MECHANICS OF ENGINEERING* 



the vertices intervening between the given segments, 
the resultant of P x P 2 P 3 acts through d. 



Here, 



329. Vertical Reaction of Piers, etc. — Fig. 354. Given the 
vertical forces or loads P 1 P 2 and P 3 acting on a rigid body 
(beam, or truss) which is supported by two piers having 
smooth horizontal surfaces (so that the reactions must be 
vertical), required the reactions V and V n of the piers. 
For an instant suppose V and V n known ; they are in 



Vo 



b^ 





c 




__„--'" 


~^-^a 










T> 


p j^~?± 


X N 


r" — { 




^-^ 


\ " 


"""j 


x— r 



V* 



A, 




Fig. 354. 

equil. with P 1 P 2 and P 3 . The introduction of the equal 
and opposite forces P and P' in the same line will not dis- 
turb the equilibrium. Taking the seven forces in the 
order P V P x P 2 P 3 V n andP', a force polygon formed with 
them will close (see (b) in Fig. where the forces which 
really lie on the same line are slightly separated). "With 
0, the butt of P, as a pole, draw the rays of the force dia- 
gram OA, OB, etc. The corresponding equil. polygon 
begins at a, the intersection of P and V in (a) (the space 
diagram), and ends at n the intersection of P' and V n . 
Join an. Now since P and P f act in the same line, an 
must be that line and must be || to P and P' of the force 
diagram. Since the amount and direction of P and P' are 
arbitrary, the position of the pole is arbitrary, while 
P u P 2 , and P 3 are the only forces known in advance in the 
force diagram. 

Hence V Q and V n may be determined as follows : Lay off 
the given loads P x , P 2 , etc., in the order of their occur- 
rence in the space diagram, to form a " load-line " AD 



GRAPHICAL STATICS. 



405 



(see (&.) Fig. 354) as a beginning for a force-diagram ; take 
any convenient pole 0, draw the rays OA, OB, 00 and 
0D. Then beginning at any convenient point a in the 
vertical line containing the unknown V , draw ab || to OA, 
be || to OB, and so on, until the last segment (dn in this 
case) cuts the vertical containing the unknown V n in some 
point ft. Join an (this is sometimes called a closing line) 
and draw a || to it through 0, in the force-diagram. This 
last line will cut the " load-line " in some point n' ', and 
divide it in two parts n' A and Dn', which are respectively 
V Q and V Q required. 

Corollary. — Evidently, for a given system of loads, in given 
vertical lines of action, and for two given piers, or abut- 
ments, having smooth horizontal surfaces, the location of the 
point ft' on the load line is independent of the choice of a 
pole. 

Of course, in treating the stresses and deflection of the 
rigid body concerned, P and P' are left out of account, as 
being imaginary and serving only a temporary purpose. 

330. Application of Foregoing Principles to a Roof Truss- 
Fig. 355. W x and W 2 are wind pressures, P l and P 2 are 
loads, while the remaining external forces, viz., the re- 




406 MECHANICS OF ENGINEERING. 

actions, or supporting forces, V , V a and H n , may be found 
by preceding §§. (We here suppose that the right abut- 
ment furnishes all the horizontal resistance ; none at the 
left). 

Lay off the forces (known) W[, W 2 , Pi, and P 2 in the 
usual way, to form a portion of the closed force polygon. 
To close the polygon it is evident we need only draw a 
horizontal through 5 and limit it by a vertical through 1. 
This determines H u but it remains to determine n' the 
point of division between V and V n . Select a convenient 
pole l9 and draw rays from it to 1, 2, etc. Assume a con- 
venient point a in the line of V in the space diagram, and 
through it draw a line || to X 1 to meet the line of W Y in 
some point b ; then a line || to 0$ to meet the line of W 2 
in some point c ; then through c || to 0$ to meet the line 
of P Y in some point d ; then through d || to X 4 to meet the 
line of P 2 in some point e, (e is identical with d, since P x 
and P 2 are in the same line) ; then ef || to X 5 to meet H n 
in some point/; then fg || to Ofi to meet V n in some 
point g. 

abcdefg is an equilibrium polygon corresponding to the 
pole Y . 

Now join ag, the " closing-line," and draw a || to it 
through Ox to determine n' t the required point of division 
between V and V n on the vertical 1 6. Hence V and V n 
are now determined as well as H Q . 

[The use of the arbitrary pole X implies the temporary 
employment of a pair of opposite and equal forces in the line 
ag, the amount of either being = 0^']. 

Having now all the external forces acting on the truss, 
and assuming that it contains no " redundant parts," i.e., 
parts unnecessary for rigidity of the frame-work, we proceed 
tc find the pulls and thrusts in the individual pieces, on 
the following plan. The truss being pin- connected, no 
piece extending beyond a joint, and all loads being con- 
sidered to act at joints, the action, pull or thrust, of each 
piece on the joint at either extremity will be in the direction 
of the piece, i.e., in a known direction, and the pin of each 



GKAPHICAL STATICS. 407 

joint is in equilibrium under a system of concurrent forces 
consisting of the loads (if any) at the joint and the pulls 
or thrusts exerted upon it by the pieces meeting there. 
Hence we may apply the principles of § 325 to each joint 
in turn, See Fig. 356. In constructing and interpreting 
the various force polygons, Mr. R. H Bow's convenient 
notation will be used ; this is as follows : In the space 
diagram a capital letter [ABC, etc.] is placed in each tri- 
angular cell of the truss, and also in each angular space in 
the outside outline of the truss between the external forces 
and the adjacent truss-pieces. In this way we can speak of 
the force W 1 as the force BC, of W 2 as the force CE, the 
stress in the piece a/3 as the force CD, and so on. That 
is, the stress in any one piece can be named from the 
letters in the spaces bordering its two sides. Corresponding 
to these capital letters in the spaces of the space-dia- 
gram, small letters will be used at the vertices of the closed 
force-polygons (one polygon for each joint) in such a way 
that the stress in the piece CD, for example, shall be the 
force ccl of the force polygon belonging to any joint in 
which that piece terminates ; the stress in the piece FG 
by the force fg in the proper force polygon, and so on. 

In Fig. 356 the whole truss is shown free, in equili- 
brium under the external forces. To find the pulls or 
thrusts (i.e., tensions or compressions) in the pieces, con- 
sider that if all but two of the forces of a closed force 
polygon are known in magnitude and direction, while the 
directions, only, of those two are known, the ivhole force 
polygon may be drawn, thus determining the amounts of 
those two forces by the lengths of the corresponding 
sides. 

We must .*. begin with a joint where no more than two 
pieces meet, as at a ; [call the joints a, /9, y, d, and the cor- 
corresponding force polygons «', /?' etc. Fig. 356.] Hence 
at a' (anywhere on the paper) make ab || and = (by scale) 
to the known force AB (i.e., V ) pointing it at the upper end, 
and from this end draw be — and || to the known force BC 
(i.e., Wi) pointing this at the lower end. 



408 



MECHANICS OF ENGINEERING. 




Fig. 356. 

To close the polygon draw through c a || to the piece 
CD, and through a a || to AD ; their intersection deter- 
mines d, and the polygon is closed. Since the arrows 
must be point to butt round the periphery, the force with 
which the piece CD acts on the pin of the joint a is a 
force of an amount = cd and in a direction from c toward 
d; hence the piece CD is in compression; whereas the 
action of the piece DA upon the pin at a is from d toward 
a (direction of arrow) and hence DA is in tension. Notice 
that in constructing the force polygon a' a right-handed 
(or clock-wise) rotation has been observed in considering 
in turn the spaces ABC and D, round the joint «. A 
similar order will be found convenient in each of the other 
joints. 

Knowing now the stress in the piece CD, (as well as in 
DA) all but two of the forces acting on the pin at the joint 
/? are known, and accordingly we begin a force polygon, /?', 
for that joint by drawing dc,= and || to the dc of polygon 
a! , but pointed in the opposite direction, since the action of 
CD on the joint j3 is equal and ojDposite to its action on 
the joint a (this disregards the weight of the piece). 
Through c draw ce = and || to the force CE (i.e., W 2 ) and 



GRAPHICAL STATICS. 



409 



pointing the same way ; then ef, = and || to the load EF 
(i.e. Pi) and pointing downward. Through f draw a || to 
the piece FG and through d, a || to the piece GD, and the 
polygon is closed, thus determining the stresses in the 
pieces FG and GD. Noting the pointing of the arrows, 
we readily see that FG is in compression while GD is in 
tension. 

Next pass to the joint d, and construct the polygon d', 
thus determining the stress gh in GH and that ad in AD ; 
this last force ad should check with its equal and oppo- 
site ad already determined in polygon a'. Another check 
consists in the proper closing of the polygon f, all of 
whose sides are now known. 

[A compound stress -diagram may be formed by super- 
posing the polygons already found in such a way as to 
make equal sides co-incide ; but the character of each 
stress is not so readily perceived then as when they are 
kept separate]. 

In a similar manner we may find the stresses in any pin- 
connected frame-work (in one plane and having no redun- 
dant pieces) under given loads, provided all the support- 
ing forces or reactions can be found. In the case of a 

braced-arch (truss) as 
shown in Fig. 357, hinged 
to the abutments at both 
ends and not free to slide 
laterally upon them, the 
reactions at and B de- 
fig. 357. pend, in amount and direc- 

tion, not only upon the equations of Statics, but on the 
form and elasticity of the arch-truss. Such cases will be 
treated later under arch -ribs, or curved beams. 




332. The Special Equil. Polygon. Its Relation to the Stresses 
in the Rigid Body. — Eeproducing Figs. 350 and 351 in Figs. 
358 and 359, (where a rigid curved beam is in equilibrium 
under the forces P lt P 2 , P 3 , P 4 and P' c ) we call a . . b . , o 



410 



MECHANICS OF ENGINEERING. 



the special equil. polygon because it corresponds to a force 
diagram in which the same order of forces has been ob- 
served as that in which they occur along the beam (from 
left to right here). From the relations between the force 



* c 

- — T\ 

A POUY.GO^ / \ 




SPACE DIAGRAM 
Fig. 358. 



FORCE DIAGRAM 
T ig, 359. 



diagram and equil. polygon, this special equil. polygon in 
the space diagram has the following properties in connec- 
tion with the corresponding rays (dotted lines) in the force 
diagram. 

The stresses in any cross-section of the portion 0' A of 
the beam, are due to P l alone ; those of any cross-section 
on AB to P l and P 2 , i.e., to their resultant P a , whose mag- 
nitude is given by the line Oa' in the force diagram, while 
its line of action is ab the first segment of the equil. poly- 
gon. Similarly, the stresses in BG are due to P l5 P 2 and 
P 3 , i.e., to their resultant P b acting along the segment be, 
its magnitude being = Ob' in the force diagram. E.g., if 
the section at m be exposed, considering O'ABm as a free 
body, we have (see Fig. 360) the elastic stresses (or inter- 



6- 





Fig. 361. 



Fig. 360. 

nal forces) at m balancing the exterior or " applied forces '* 
and P 3 . Obviously, then, the stresses at m are just 



P»P: 



GRAPHICAL STATICS. 411 

the same as if B h the resultant of P l9 P 2 and P 3 , acted upon 
an imaginary rigid prolongation of the beam intersecting 
be (see Fig. 361). B h might be called the " anti-stress -result- 
ant " for the portion BG of the beam. We may .*. state 
the following : If a rigid body is in equilibrium under a sys- 
tem of Non-Concurrent Forces in a plane, and the special equi- 
librium polygon lias been drawn, then each ray of the force 
diagram is the anti-stress-resultant of that portion of the beam 
which corresponds to the segment of the equilibrium polygon 
to which the ray is parallel ; and its line of action is the seg- 
ment just mentioned. 

Evidently if the body is not one rigid piece, but com- 
posed of a ring of uncemented blocks (or voussoirs), it may 
be considered rigid only so long as no slipping takes place 
or disarrangement of the blocks ; and this requires that the 
" anti-stress-resultant " for a given joint between two 
blocks shall not lie outside the bearing surface of the 
joint, nor make too small an angle with it, lest tipping or 
slipping occur. For an example of this see Fig. 362, show- 
ing a line of three blocks in equilibrium under five forces. 

The pressure borne at the 
^ joint MN, is = i? a in the 
| p force -diagram and acts in 
the line ab. The con- 
struction supposes all 
the forces given except 
fig. 362. one, in amount and posi- 

tion, and that this one could easily be found in amount, as 
being the side remaining to close the force polygon, while 
its position would depend on the equil. polygon. But in 
practice the two forces P 1 and B' c are generally unknown, 
hence the point 0, or pole of the force diagram, can not 
be fixed, nor the special equil. polygon located, until other 
considerations, outside of those so far .presented, are 
brought into play. In the progress of such a problem, as 
will be seen, it will be necessary to use arbitrary trial po- 
sitions for the pole 0, and corresponding trial equilibrium 
polygons. 




412 



MECHANICS OF ESGIXEEIilNG. 



CHAPTEE IX. 



GRAPHICAL STATICS OF VERTICAL FORCES, 



333. Remarks. — (With the exception of § 378 a) in prob= 
lems to be treated subsequently (either the stiff arch-rib ? 
or the block-work of an arch-ring, of masonry) when the 
body is considered free all the forces holding it in equil. 
will be vertical (loads, due to gravity) except the reactions 
at the two extremities, as in Fig. 363 ; but for convenience 
each reaction will be replaced by its horizontal and verti- 
cal components (see Fig. 364). The two ZTs are of course 
equal, since they are the only horizontal forces in the 
system. Henceforth, all equil. polygons under discussion will 
be understood to imply this kind of system of forces. P 1} i\, 



rP T 2 |P 3 Ki \ 



* P P 'v, 




Fig. 363. 



FIG. 364. 




h 



P 3 >, 



FIG. 364a. 



etc., will represent the " loads " ; V and V n the vertical 
components of the abutment reactions ; H the value of 
either horizontal component of the same. (We here sup- 
pose the pressures T and T n resolved along the horizon- 
tal and vertical.) 



GRAPHICAL STATICS. 



413 



334. Concrete Conception of an Equilibrium Polygon, — Any 
equilibrium polygon has this property, due to its mode 
of construction, viz.: If the ab and be of Fig. 358 were im- 
ponderable straight rods, jointed at b without friction, they 
would be in equilibrium under the system of forces there 
given. (See Fig. 364a). The rod ab suffers a compression 
equal to the P a of the force diagram, Fig. 359, and be a 
compression = B h . In some cases these rods might be in 
tension, and would then form a set of links playing the 
part of a suspension-bridge cable. (See § 44). 

335. Example of Equilibrium Polygon Drawn to Vertical Loads 
— Fig. 365. [The structure bearing the given loads is not 
shown, but simply the imaginary rods, or segments of an 
equilibrium polygon, which would support the given loads 
in equilibrium if the abutment points A and B, to which 
the terminal rods are hinged, were firm. In the present 
case this equilibrium is unstable since the rods form a 
standing structure ; but if they were hanging, the equilibri- 
um would be stable. Still, in the present case, a very light 
bracing, or a little friction at all joints would make the 
equilibrium stable. 




2. FT. TO INCH 



/ 

/v. 

/ X"'n 

/££>* - — - — ' 

)#-" " *. 


-*^ . 

800LBS. TojNc ^^yn 



14 



Fig. 365. 



Given three loads P lt P 2 , and P 3 , and two " abutment 
verticals " A' and B\ in which we desire the equil. poly- 
gon to terminate, lay off as a "load-line" to scale, P lf P 2 , 
and P 3 end to end in their order. Then selecting any pole, 



414 MECHANICS OF ENGINEERING. 

0, draw the rays 01, 02, etc., of a force diagram (the V's 
and P's, though really on the same vertical, are separated 
slightly for distinctness ; also the iPs, which both pass 
through and divide the load-line into V and V n ). We 
determine a corresponding equilibrium polygon by draw- 
ing through A (any point in A) a line || to . . 1, to inter- 
sect P l in some point b ; through b a || to . . 2, and so- on* 
until B' the other abutment-vertical is struck in some 
point B. AB is the " abutment-line " or " closing -line." 

By choosing another point for 0, another equilibrium 
polygon would result. As to which of the infinite 
number (which could thus be drawn, for the given loads 
and the A and B' verticals) is the special equilibrium poly- 
gon for the arch-rib or stone-arch, or other structure, on 
which the loads rest, is to be considered hereafter. In 
any of the above equilibrium polygons the imaginary 
series of jointed rods would be in equilibrium. 

336, Useful Property of an Equilibrium Polygon for Vertical 
Loads. — (Particular case of § 328). See Fig. 366. In any 
equil. polygon, supporting vertical loads, consider as free 
any number of consecutive segments, or rods, with the 
loads at their joints, e. g., the 5th and 6th and portions of 
C/jv N the 4th and 7th which, we sup- 

/ 5 >viX^ pose cut and the compressive 

— ~Hs^\ forces in them put in, T\ and 
\ T 7 , in order to consider 4 5 6 7 
v \ * as a free body. For equil., 

"""*T\ according to Statics, the lines 

5 lp e \ of action of T, and T 7 (the com- 

Fig. 366. pression in those rods) must in- 

tersect in a point, (7, in the line of action of the resultant 
of P 4 , P 5 , and P 6 ; i.e., of the loads occurring at the inter- 
vening vertices. That is, the point C must lie in the ver- 
tical containing the centre of gravity of those loads. Since 
the position of this vertical must be independent of the 
particular equilibrium polygon used, any other (dotted 
lines in Fig. 366) for the same loads will give the same re- 



GRAPHICAL STATICS. 415 

suits. Hence the vertical CD, containing the centre of 
gravity of any number of consecutive loads, is easily found 
by drawing the equilibrium polygon corresponding to 
any convenient force diagram having the proper load-line. 
This principle can be advantageously applied to finding 
a gravity -line of any plane figure, by dividing the latter 
into parallel strips, whose areas may be treated as loads 
applied in their respective centres of gravity. If the strips 
are quite numerous, the centre of gravity of each may be 
considered to be at the centre of the line joining the mid- 
dles of the two long sides, while their areas may be taken 
as proportional to the lengths of the lines drawn through 
these centres of gravity parallel to the long sides and lim- 
ited by the end-curves of the strips. Hence the " load- 
line " of the force diagram may consist of these lines, or 
of their halves, or quarters, etc., if more convenient (§ 376). 



USEFUL RELATIONS BETWEEN FORCE DIA- 
GRAMS AND EQUILIBRIUM POLYGONS, 
(for vertical loads.) 

337. R6sum6 of Construction. — Fig. 367. Given the loads 
P u etc., their verticals, and the two abutment verticals A' 
and B', in which the abutments are to lie ; we lay off a 
load-line 1 ... 4, take any convenient pole, 0, for a force- 
diagram and complete the latter. For a corresponding 
equilibrium polygon, assume any point A in the vertical 
A\ for an abutment, and draw the successive segments 
Al, 2, etc., respectively parallel to the inclined lines of the 
force diagram (rays), thus determining finally the abut- 
ment B, in B\ which (B) will not in general lie in the hor- 
izontal through A. 

Now join AB, calling AB the abutment-line, and draw a 
parallel to it through 0, thus fixing the point nf on the 



416 



MECHANICS OF ENGINEERING. 



A ! 

1 ft 


\ ft 

3 1 

1^4 




1 

2 


/ 

to 


Pi 


P 2 


P 3 




1/1 


H * i ^ 




n 

3 r 














5 


i 


£ 


\ H l 




4^ 





Fig. 367. 



Fig. 368. 



load-line. This point w', as above determined, is indepen- 
dent of the location of the pole, 0, (proved in § 329) and 
divides the load-line into two portions (V' = 1 . . . n', and 
V' a = n' : . . 4) which are the vertical pressures which two 
supports in the verticals A' and B' ivould sustain if the 
given loads rested on a horizontal rigid bar, as in Fig. 368. 

See § 329. Hence to find the point n' we may use any 
convenient pole 0. 

[N. B.— The forces V and V n of Fig. 367 are not identi- 
cal with V' and V n9 but may be obtained by dropping a 
"| from to the load-line, thus dividing the load-line 
into two portions which are V (upper portion) and V n . 
However, if A and B be connected by a tie-rod, in Fig. 
367, the abutments in that figure will bear vertical press- 
ures only and they will be the same as in Fig. 368, while 
the tension in the tie-rod will be ==. On'.'] 



338. Theorem. — The vertical dimensions of any tivo equili- 
brium polygons, draivn to the same loads, load-verticals, and 
abutment-verticals, are inversely proportioned to their IT 's (or 
"pole distances "). We here regard an equil. polygon and 
its abutment-line as a closed figure. Thus, in Fig. 369, 
we have two force-diagrams (with a common load-line, for 
convenience) and their corresponding equil. polygons, for 
the same loads and verticals. From § 337 we know that 
On' is || to AB and n' is II to A B . Let CD be any ver- 
tical cutting the first segments of the two equil. polygons. 



GRAPHICAL STATICS. 



417 



Denote the intercepts thus determined by z' and z' oi respect- 
} f r ively. From the 

| I Ho 

Ibi 



parallelisms just 
mentioned, and 
others more famil- 
^ iar, we have the 
triangle In' sim- 
ilar to the triangle 
Az' (shaded), and 
the triangle O ln' 
similar to the tri- 
angle A z£. Hence 




the proportions between j In' z 



bases and altitudes 



= - and 



In' 



{ H h H h 

.'. z' : z' : : H : H. The same kind of proof may easily 

be applied to the vertical intercepts in any other segments, 
e. g., z" and z" . Q. E. D. 

339. Corollaries to the foregoing. It is evident that : 
(1.) If the pole of the force-diagram be moved along a 
vertical line, the equilibrium polygon changing its form 
in a corresponding manner, the vertical dimensions of the 
equilibrium polygon remain unchanged ; and 

(2.) If the pole move along a straight line which con- 
tains the point n', the direction of the abutment-line 
remains constantly parallel to the former line, while the 
vertical dimensions of the equilibrium polygon change in 
inverse proportion to the pole distance, or H, of the force- 
diagram. [If is the "1 distance of the pole from the load- 
line, and is called the pole-distance]. 

§ 340. Linear Arch as Equilibrium Polygon. — (See § 316.) 
If the given loads are infinitely small with infinitely small 
horizontal spaces between them, any equilibrium polygon 
becomes a linear arch. Graphically we can not deal with 
these infinitely small loads and spaces, but from § 336 it 
is evident that if we replace them, in successive groups. 



418 



MECHANICS OF ENGINEERING. 




Fig. 370. 



by finite forces, each of which == the sum of those com- 
posing one group and is 
applied through the cen- 
tre of gravity of that 
group, we can draw an 
equilibrium polygon 
whose segments will be 
tangent to the curve of 
the corresponding linear 
arch, and indicate its posi- 
tion with sufficient exactness for practical purposes. (See 
Fig. 370). The successive points of tangency A, m, n, etc., 
lie vertically under the points of division between the 
groups. This relation forms the basis of the graphical 
treatment of voussoir, or blockwork, arches. 

341. To Pass an Equilibrium Polygon Through Three Arbitrary 
Points. — (In the present case the forces are vertical. For 
a construction dealing with any plane system of forces see 
construction in § 378$.) Given a system of loads, it is re- 

j quired to draw 

I* y ~\ an equilibrium 

■„ polygon for 

them through 

;~P^ [ g anythree points, 

two of which 
4 may be consid- 
ered as abut- 
ments, outside of the load-verticals, the third point being 
between the verticals of the first two. See Fig. 371. The 
loads P lf etc., are given, with their verticals, while A, p, 
and B are the three points. Lay off the load-line, and 
with any convenient pole, Y , construct a force-diagram, 
then a corresponding preliminary equilibrium polygon 
beginning at A. Its right abutment B u in the vertical 
through B, is thus found. Y n f can now be drawn || to AB X , 
to determine n'. Draw n'O II to BA. The pole of the 
required equilibrium polygon must lie on n'O (§ 337} 




Fig. 371. 



GRAPHICAL STATICS. 419 

Draw a vertical through p. The H of the required equili- 
brium polygon must satisfy the proportion H : H x : : rs : 
pm. (See § 338). Hence construct or compute H from 
the proportion and draw a vertical at distance H from 
the load-line (on the left of the load-line here) ; its inter- 
section with n' gives the desired pole, for which a 
force diagram may now be drawn. The corresponding 
equilibrium polygon beginning at the first point A will 
also pass through p and B ; it is not drawn in the figure. 

342. Symmetrical Case of the Foregoing Problem.— If two 
points A and B are on a level, the third, p, on the middle 
vertical between them ; and the loads (an even number) 
symmetrically disposed both in position and magnitude, about 
p, we may proceed more simply, as follows : (Fig. 372). 

From symmetry n f 
// must occur in the mid- 
■ die of the load-line, of 
which we need lay off 
only the upper half. 
Take a convenient pole 
FlG . 372. Oi, in the horizontal 

through n', and draw a half force diagram and a corres- 
ponding half equilibrium polygon (both dotted). The up- 
per segment be of the latter must be horizontal and being 
prolonged, cuts the prolongation of the first segment in a 
point d, which determines the vertical CD containing the 
centre of gravity of the loads occurring over the half -span 
on the left. (See § 336). In the required equilibrium poly- 
gon the segment containing the point p must be horizon- 
tal, and its intersection with the first segment must lie in 
CD. Hence determine this intersection, C, by drawing the 
vertical CD and a horizontal through p ; then join CA, 
which is the first segment of the required equil. polygon. 
A parallel to CA through 1 is the first ray of the corres- 
ponding force diagram, and determines the pole on the 
horizontal through n\ Completing the force diagram for 



»B 



420 



MECHANICS OF ENGINEERING. 



this pole (half of it only here), the required equil. poly- 
gon is easily finished afterwards. 

343. To Find a System of Loads Under Which a Given Equi- 
librium Polygon Would be in Equilibrium, — Fig. 373. Let AB 
be the given equilibrium polygon. Through any point 
as a pole draw a parallel to each 
segment of the equilibrium polygon. 
Any vertical, as V, cutting these 
lines will have, intercepted upon it, 
a load-line 1, 2, 3, whose parts 1 . . 2, 
2 . . 3, etc., are proportional to the 
successive loads which, placed on 
the corresponding joints of the equilibrium polygon would 
be supported by it in equilibrium (unstable). 

One load may be assumed and the others constructed. 
A hanging, as well as a standing, equilibrium polygon 
may be dealt with m like manner, but will be instable equi- 
librium. The problem in § 44 may be solved in this way„ 




Fig. 373. 



AECHES OF MASONKY. 



421 



CHAPTER X. 



RIGHT ARCHES OF MASONRY. 



344. — In an ordinary " right " stone-arch (i.e., one in 
which the faces are ~| to the axis of the cylindrical soffit, 
or under surface), the successive blocks forming the arch- 
ring are called voussoirs, the joints between them being 
planes which, prolonged, meet generally in one or more 
horizontal lines ; e.g., those of a three-centred arch in three 
j| horizontal lines ; those of a circular arch in one, the axis 
of the cylinder, etc. Elliptic arches are sometimes used. The 
inner concave surface is called the soffit, to which the radiat- 
ing joints between the voussoirs are made perpendicular. 
The curved line in which the soffit is intersected by a plane 




H to the axis of the arch is the Intrados. The curve in the 
same plane as the intrados, and bounding the outer ex- 
tremities of the joints between the voussoirs, is called the 
Extrados. 

Fig. 374 gives other terms in use in connection with a 



422 



MECHANICS OF ENGINEEKLNG. 



stone arch, and explains those already given. AB is the 
" springing-line." 

345o Mortar and Friction. — As common mortar hardens 
very slowly, no reliance should be placed on its tenacity 
as an element of stability in arches of any considerable 
size ; though hydraulic mortar and thin joints of ordinary 
mortar can sometimes be depended on. Friction, however, 
between the surfaces of contiguous voussoirs, plays an 
essential part in the stability of an arch, and will there- 
fore be considered. 

The stability of voussoir-arches must .\ be made to 
depend on the resistance of the voussoirs to compresssion 
and to sliding upon each other ; as also of the blocks 
composing the piers, the foundations of the latter being 
firm. 

346. Point of Application of the Resultant Pressure between 
two consecutive voussoirs ; (or pier blocks). Applying 
Navier's principle (as in flexure of beams) that the press- 
ure per unit area on a joint varies uniformly from the 
extremity under greatest compression to the point of least 
compression (or of no compression); and remembering 
that negative pressures (i.e., tension) can not exist, as they 
might in a curved beam, we may represent the pressure 
per unit area at successive points of a joint (from the intra- 
dos toward the extrados, or vice versa.) by the ordinates of 
a straight line, forming the surface of a trapezoid or tri- 
angle, in which figure the foot of the ordinate of the cen- 
tre of gravity is the point of application of the resultant 
pressure. Thus, where the least compression is supposed 




*„^0f b/ 1 



Fig. 375. 





Fig. 3 1 



Fig. 377. 



Fig. 378. 



MASONRY ARCHES. 



423 



to occur at the intrados A, Fig. 375, the pressures vary as 
the ordinates of a trapezoid, increasing to a maximum value 
at B, in the extrados. In Fig. 376, where the pressure is zero 
at B, and varies as the ordinates of a triangle, the result- 
ant pressure acts through a point one-third the joint- 
length from A. Similarly in Fig. 377, it acts one-third 
the joint-length from B. Hence, when the pressure is not 
zero at either edge the resultant pressure acts within the 
middle third of the joint. Whereas, if the resultant press- 
ure falls without the middle third, it shows that a portion 
Am of the joint, see Fig. 378, receives no pressure, i.e., the 
joint tends to open along Am. 

Therefore that no joint tend to open, the resultant press- 
ure must fall within the middle third. 

It must be understood that the joint surfaces here dealt 
with are rectangles, seen edgewise in the figures. 

347. Friction. — By experiment it has been found the 
angle of friction (see § 156) for two contiguous voussoirs 
of stone or brick is about 30° ; i.e., the coefficient of fric- 
tion is / = tan. 30°. Hence if the direction of the press- 
ure exerted upon a voussoir by its neighbor makes an 
angle a less than 30° with the normal to the joint surface, 
there is no danger of rupture of the arch by the sliding 
of one on the other. (See Fig. 379). 



348. Resistance to Crushing. — When the resultant pressure 
falls at its extreme allowable limit, viz. : the edge of the 
middle third, the pressure per 
unit of area at n, Fig. 380, is 
double the mean pressure per 
unit of area. Hence, in de- 
signing an arch of masonry, 
we must be assured that at 
every joint (taking 10 as a 
factor of safety) 




Fig. 37 



j Double the mean press- 
j ure per unit of area 



l: 



must be less than 1 / l0 G 



424 MECHANICS OF ENGINEERING. 

C being the ultimate resistance to crushing, of the material 
employed (§ 201) (Modulus of Crushing). 

Since a lamina one foot thick will always be considered 
in what follows, careful attention must be paid to the units 
employed in applying the above tests. 

Example. — If a joint is 3 ft. by 1 foot, and the resultant 
pressure is 22.5 tons the mean pressure per sq. foot is 

£>=22.5-f-3=7.5 tons per sq. foot 

,\ its double=15 tons per sq. foot=208.3 lbs. sq. inch, 
which is much less than 1 / w of C for most building stones ; 
see § 203, and below. 

At joints where the resultant pressure falls at the middle, 
the max. pressure per square inch would be equal to the 
mean pressure per square inch ; but for safety it is best to 
assume that, at times, (from moving loads, or vibrations) 
it may move to the edge of the middle third, causing the 
max. pressure to be double the mean (per square inch). 

Gen. Gillmore's experiments in 1876 gave the following 
results, among, many others : 

NAME OF BUILDING STONE. C IN LBS. PER SQ. INCH. 

Berea sand-stone, 2-inch cube, - 8955 

4l " " - - 11720 

Limestone, Sebastopol, 2-inch cube (chalk), - - 1075 

Limestone from Caen, France, - 3650 

Limestone from Kingston, N. Y., - 13900 
Marble, Vermont, 2-inch cube, - - 8000 to 13000 

Granite, New Hampshire, 2-inch cube, 15700 to 24000 

349. The Three Conditions of Safe Equilibrium for an arch of 
uncemented voussoirs. 

■ Recapitulating the results of the foregoing paragraphs, 
we may state, as follows, the three conditions which must 
be satisfied at every joint of arch-ring and pier, for each 
of any possible combination of loads upon the structure : 

(1). The resultant pressure must pass within the middle- 
third 

(2). The resultant pressure must not make an angle > 
30° with the normal to the joint. 

(3). The m^an pressure per unit of area on the surface 



ARCH OF MASONRY. 



425 




of the joint must not exceed 1 / 20 of the Modulus of crush- 
ing of the material. 

350. The True Linear-Arch, or Special Equilibrium Polygon ; 

and the resultant pressure at any joint. Let the weight 
of each voussoir and its load be represented by a vertical 
force passing through the centre of gravity of the two, as 
in Fig. 381. Taking any 
two points A and B, A 
being in the first joint and 
B in the last ; also a third 
point, p, in the crown 
joint (supposing such to 
be there, although gener- 
ally a key-stone occupies | 
the crown), through these Fig. 3si. 

three points can be drawn [§ 341] an equilibrium polygon 
for the loads given ; suppose this equil. polygon nowhere 
passes outside of the arch-ring (the arch-ring is the por- 
tion between the intrados, mw, and the (dotted) extrados 
m'n') intersecting the joints at b, c, etc. Evidently if such 
be the case, and small metal rods (not round) were insert- 
ed at A, b, c, etc., so as to separate the arch-stones slight- 
ly, the arch would stand, though in unstable equilibrium, 
the piers being firm ; and by a different choice of A, p, and 
B, it might be possible to draw other equilibrium poly- 
gons with segments cutting the joints within the arch- 
ring, and if the metal rods were shifted to these new inter- 
sections the arch would again stand (in unstable equilib- 
rium). 

In other words, if an arch stands, it may be possible to 
draw a great number of linear arches within the limits of 
the arch -ring, since three points determine an equilibrium 
polygon (or linear arch) for given loads. The question 
arises then : which linear arch is the locus of the actual re- 
sultant pressures at the successive joints ? 

[Considering the arch-ring as an elastic curved beam 
inserted in firm piers (i.e., the blocks at the springing-line 



426 MECHANICS OF ENGINEERING. 

are incapable of turning) and having secured a close fit at 
all joints before the centering is lowered, the most satisfac- 
tory answer to this question is given in Prof. Greene's 
" Arches," p. 131 ; viz., to consider the arch-ring as an 
arch rib of fixed ends and no hinges ; see § 380 of next 
chapter ; but the lengthy computations there employed 
(and the method demands a simple algebraic curve for the 
arch) may be most advantageously replaced by Prof. 
Eddy's graphic method (" New Constructions in Graphical 
Statics," published in Van Nostrand's Magazine for 1877), 
which applies to arch curves of any form. 

This method will be given in a subsequent chapter, on 
Arch Bibs, or Curved Beams ; but for arches of masonry a 
much simpler procedure is sufficiently exact for practical 
purposes and will now be presented]. 

If two elastic blocks 
of an arch-ring touch at 
one edge, Fig. 382, their 
adjacent sides making a 
small angle with each 
f ig- 3S2 - FlG - 3S3 - other, and are then grad- 

ually pressed more and more forcibly together at the edge 
m, as the arch-ring settles, the centering being gradually 
lowered, the surface of contact becomes larger and larger, 
from the compression which ensues (see Fig. 383); while 
the resultant pressure between the blocks, first applied at 
the extreme edge m, has now probably advanced nearer the 
middle of the joint in the mutual adjustment of the arch - 
stones. With this in view we may reasonably deduce the 
following theory of the location of the true linear arch 
(sometimes called the " line of pressures " and " curve of 
pressure") in an arch under given loading and with firm 
'piers. (Whether the piers are really unyielding, under the 
oblique thrusts at the springing-line, is a matter for sub- 
sequent investigation. 

351. Location of the True Linear Arch. — Granted that the 
voussoirs have been closely fitted to each other over the 





ARCH OF MASONRY. 427 

centering (sheets of lead are sometimes used in the joints 
to make a better distribution of pressure); and that the 
piers are firm ; and that the arch can stand at all without 
the centering ; then we assume that in the mutual accom- 
modation between the voussoirs, as the centering is low- 
ered, the resultant of the pressures distributed over any 
joint, if at first near the extreme edge of the joint, advances 
nearer to the middle as the arch settles to its final posi- 
tion of equilibrium under its load ; and hence the follow^ 
ing 

352. Practical Conclusions. 

I. If for a given arch and loading, with firm piers, an 
equilibrium polygon can be drawn (by proper selection of 
the points A, p, and B, Fig. 381) entirely within the mid- 
dle third of the arch ring, not only will the arch stand, but 
the resultant pressure at every joint will be within the 
middle third (Condition 1, § 349) ; and among all possible 
equilibrium polygons which can be drawn within the mid- 
dle third, that is the " true " one which most nearly coin- 
cides with the middle line of the arch-ring. 

II. If (with firm piers, as before) no equilibrium poly- 
gon can be drawn within the middle third, and only one 
within the arch-ring at all, the arch may stand, but chip- 
ping and spawling are likely to occur at the edges of the 
joints. The design should .*. be altered. 

III. If no equilibrium polygon can be drawn within 
the arch-ring, the design of either the arch or the loading 
must be changed ; since, although the arch may stand, 
from the resistance of the spandrel walls, such a stability 
must be looked upon as precarious and not countenanced 
in any large important structure. (Very frequently, in 
small arches of brick and stone, as they occur in buildings, 
the cement is so tenacious that the whole structure is vir- 
tually a single continuous mass). 

When the " true " linear arch has once been determined, 
the amount of the resultant pressure on any joint is given 
by the length of the proper ray in the force diagram. 



428 



MECHANICS OF ENGINEERING. 



ARRANGEMENT OF DATA FOR GRAPHIC 
TREATMENT. 

353. Character of Load. — In most large stone arch bridges 
the load (permanent load) does not consist exclusively of 
masonry up to the road-way but partially of earth filling 
above the masonry, except at the faces of the arch where 
the spandrel walls serve as retaining walls to hold the 
earth. (Fig, 384). If the intrados is a half circle or half- 




Fig. 384. Fig. 385. 

ellipse, a compactly -built masonry backing is carried up 
beyond the springing-line to AB about 60° to 45° from the 
crown, Fig. 385 ; so that the portion of arch ring below 
AB may be considered as part of the abutment, and thus 
AB is the virtual springing-line, for graphic treatment. 

Sometimes, to save filling, small arches are built over 
the haunches of the main arch, with earth placed over 
them, as shown in Fig. 386. In any of the preceding cases 




Fig. 38G. 



Fig. 38? 



it is customary to consider that, on account of the bond- 
ing of the stones in the arch shell, the loading at a given 
distance from the crown is uniformly distributed over the 
width of the roadway. 



ARCHES OF MASONRY. 429 

354, Reduced Load-Contour.— In the graphical discussion 
of a proposed arch we consider a lamina one foot thick, 
this lamina being vertical and T to the axis of the arch ; 
i.e., the lamina is || to the spandrel walls. For graphical 
treatment, equal areas of the elevation (see Fig. 387) of 
this lamina must represent equal weights. Taking the 
material of the arch-ring as a standard, we must find for 
each point p of the extrados an imaginary height z of the 
arch-ring material, which would give the same pressure 
(per running horizontal foot) at that point as that due to 
the actual load above that point. A number of such or- 
dinates, each measured vertically upward from the extra- 
dos determine points in the "Reduced Load-Contour," i.e., 
the imaginary line, AM, the area between which and the 
extrados of the arch -ring represents a homogeneous load 
of the same density as the arch-ring, and equivalent to the 
actual load (above extrados), vertical by vertical. 

355. Example of Reduced Load-Contour. — Fig. 388. Given 
an arch-ring of granite (heaviness = 170 lbs. per cubic 
foot) with a dead load of rubble (heav. = 140) and earth 
(heav. = 100), distributed as in figure. At the point p, of 
the extrados, the depth 5 feet of rubble is equivalent to a 
depth of [j^x5]-4.1 ft. of granite, while the 6 feet of earth 
is equivalent to [^?x6]=3.5 feet of granite. Hence the 
Reduced Load-Contour has an ordinate, above p, of 7.6 feet. 
That is, for each of several points of the arch-ring extrados 
reduce the rubble ordinate in the ratio of 170 : 140, and 
the earth ordinate in the ratio 170 : 100 and add the re- 
sults, setting off the sum vertically from the points in the 
extrados*. In this way Fig. 389 is obtained and the area 



*This is most conveniently done by graphics, thus : On a right-line set off 17 equal 
parts (of any convenient magnitude.) Call this distance OA. Through draw another 
right line at any convenient angle (30° to 60°) with OA, and on it from 

set off OB equal to 14 (for the rubble ; or 10 for the earth) of the same equal 
parts. Join AB. From toward A set off* all the rubble ordinates to be reduced, 
feach being set off from 0) and through the other extremity of each draw a line par- 
allel to AB. The reduced ordinates will be the respective lengths, from 0, along OB, 
to the intersections of these parallels with OB. 

* With the dividers. 



430 



MECHANICS OF ENGINEERING. 




there given is to be treated as representing homogeneous 
granite one foot thick. This, of course, now includes the 
arch-ring also. AB is the " reduced load-contour." 

356. Live Loads. — In discussing a railroad arch bridge 
the " live load " (a train of locomotives, e.g., to take an ex- 
treme case) can not be disregarded, and for each of its po- 
sitions we have a separate Reduced Load-Contour. 

Example. — Suppose the arch of Fig. 388 to be 12 feet 
wide (not including spandrel walls) and that a train of lo- 
comotives weighing 3,000 lbs. per running foot of the track 
covers one half of the span. Uniformly * distributed later- 
ally over the width, 12 ft., this rate of loading is equiva- 
lent to a masonry load of one foot high and a heaviness of 
250 lbs. per cubic ft., i.e., is equivalent to a height of 1.4 
ft. of granite masonry [since —■ X 1.0— 1.4] over the half 
span considered. Hence from Fig. 390 we obtain Fig. 391 
in an obvious manner. Fig. 391 is now ready for graphic 
treatment. 




Fig. 390. Fig. 391. 

357. Piers and Abutments. — In a series of equal arches 
the pier between two consecutive arches bears simply the 
weight of the two adjacent semi-arches, plus the load im- 

* If the earth-filling is shallow, the laminae directly under the track prol> 
ably receive a greater pressure than the others. 



ARCHES OF MASONKY. 



431 



mediately above the pier, and .-. doss not need to be as 
large as the abutment of the first and last arches, since 
these latter must be prepared to resist the oblique thrusts 
of their arches without help from the thrust of another on 
the other side. 

In a very long series of arches it is sometimes customary 
to make a few of the intermediate piers large enough to 
act as abutments. These are called " abutment piers," and 
in case one arch should fall, no others would be lost except 
those occurring between the same two abutment piers as 
the first. See Fig. 392. A is an abutment-pier. 



00-000 



Fig. 3'J; 



GRAPHICAL, TREATMENT OF ARCH. 

358. — Having found the " reduced load-contour," as in 
preceding paragraphs, for a given arch and load, we are 
ready to proceed with the graphic treatment, i.e., the first 
given, or assumed, form and thickness of arch-ring is to be 
investigated with regard to stability. It may be necessary 
to treat, separately, a lamina under the spandrel wall, and 
one under the interior loading. The constructions are 
equally well adapted to arches of all shapes, to Gothic as 
well as circular and elliptical. 

359.— Case I. Symmetrical Arch and Symmetrical Loading. — . 
(The " steady " (permanent) or " dead " load on an arch is 
usually symmetrical). Fig. 393. From symmetry we need 




Fig. 393. 



Fig. 394. 



Fig. 395. 



432 MECHANICS OE ENGINEEKLNG. 

deal with only one half (say the left) of the arch and load. 
Divide this semi-arch and load into six or ten divisions 
by vertical lines ; these divisions are considered as trape- 
zoids and should have the same horizontal width = b (a 
convenient whole number of feet) except the last one, LKN 9 
next the abutment, and this is a pentagon of a different 
width b l9 (the remnant of the horizontal distance LC). The 
weight of masonry in each division is equal to (the area 
of division) X (unity thickness of lamina) x (weight of a cu- 
bic unit of arch-ring). For example for a division having 
an area of 20 sq. feet, and composed of masonry weighing 
160 lbs. per cubic foot, we have 20x1x160=3,200 lbs., 
applied through the centre of gravity of the division. 
The area of a trapezoid, Fig. 394, is ^fr(Ai+A 2 )> and its cen- 
tre of gravity may be found, Fig. 395, by the construction 
of Prob. 6, in § 26 ; or by § 27a. The weight of the pen- 
tagon LN, Figo 393, and its line of application (through 
centre of gravity) may be found by combining results for 
the two trapezoids into which it is divided by a vertical 
through K. See § 21. 

Since the weights of the respective trapezoids {except, 
ing LN) are proportional to their middle vertical in- 
tercepts [such as ^2(7*1+^2) Fig. 394] these intercepts (trans- 
ferred with the dividers) may be used directly to form the 
load-line, Fig. 396, or proportional parts of them if more 
convenient. The force scale, which this implies, is easily 
computed^ and a proper length calculated to represent the 
weight of the odd division LN ; i.e., 1 ... 2 on the load- 
line. 

Now consider A, the middle point of the abutment joint, 
Fig. 396, as the starting point of an equilibrium polygon 
(or abutment of a linear arch) for a given loading, and re- 
quire that this equilibrium polygon shall pass through p, 
the middle of the crown joint, and through the middle of 
the abutment joint on the right (not shown in figure). 

Proceed as in § 342, thus determining the polygon Ap 
for the half-arch. Draw joints in the arch-ring through 
those points where the extrados is intersected by the ver- 






ARCHES OF MASONRY. 



433 




Fig. 390 



ileal separating the divisions (not the gravity verticals). 
The points in which these joints are cut by the segments 
of the equilibrium polygon, Fig. 397, are (very nearly, if 
the joint is not more than 60° from p t the crown) the points 
of application in these joints, respectively, of the resultant 
pressures on them, (if this is the " true linear arch " for 
this arch and load) while the amount and direction of each 
such pressure is given by the proper ray in the force-dia- 
gram. 

If at any joint so drawn the linear arch (or equilibrium 
polygon) passes outside the middle third of the arch-ring, 
the point A, or p, (or both) should be judiciously moved 
(within the middle third) to find if possible a linear arch 
which keeps within limits at all joints. If this is found 
impossible, the thickness of the arch -ring may be increased 
at the abutment (giving a smaller increase toward the 
crown) and the desired result obtained ; or a change in the 
distribution or amount of the loading, if allowable, may 
gain this object. If but one linear arch can be drawn 
within the middle third, it may be considered the " true " 
one ; if several, the one most nearly co-inciding with the 
middles of the joints (see §§ 351 and 352) is so considered. 



360.— Case II. TJnsymmetrical Loading on a Symmetrical Arch; 
(e.g., arch with live load covering one half -span as in Figs. 
390 and 391). Here we must evidently use a full force 
diagram, and the full elevation of the arch -ring and load. 



434 



MECHANICS OF 'ENGINEERING. 



See Fig. 398. Select three points A, p, and B, as follows, 
to determine a trial equilibrium polygon : 

Select A at the lower limit of the middle third of the 




abutment-joint at the end of the span which is the more 
heavily -loaded ; in the other abutment-joint take B at the 
upper limit of the middle third ; and take p in the middle 
of the crown-joint. Then by § 341 draw an equilibrium 
polygon (i.e., a linear arch) through these three points for 
the given set of loads, and if it does not remain within the 
middle third, try other positions for A, p, and B, within 
the middle third. As to the " true linear arch " alterations 
of the design, etc., the same remarks apply as already 
given in Case I. Yery frequently it is not necessary to 
draw more than one linear arch, for a given loading, for 
even if one could be drawn nearer the middle of the arch- 
ring than the first, that fact is most always apparent on 
mere inspection, and the one already drawn (if within 
middle third) will furnish values sufficiently accurate for 
the pressures on the respective joints, and their direction 
angles. 

360a. — The design for the arch-ring and loading is not 
to be considered satisfactory until it is ascertained that for 
the dead load and any possible combination of live-load 
(in addition) the pressure at any joint is 



ARCHES OF MASONRY. 



435 



(1.) Within the middle third of that joint ; 

(2.) At an angle of < 30° with the normal to joint- 
surface. 

(3.) Of a mean pressure per square inch not > than 1 / 20 
of the ultimate crushing resistance. (See § 348.) 

§ 361. Abutments. — The abutment should be compactly 
and solidly built, and is then treated as a single rigid mass. 
The pressure of the lowest voussoir upon it (considering 
a lamina one foot thick) is given by the proper ray of the 
force diagram (0 .. 1, e. g., in Fig. 396) in amount and direc- 
tion. The stability of the abutment will depend on the 
amount and direction of the resultant obtained by com- 
bining that pressure P a with the weight G of the abutment 
and its load, see Fig. 399. Assume a probable width RS 
for the abutment and compute the weight G 
of the corresponding abutment OBRS and 
MNBO, and find the centre of gravity of the 
whole mass C. Apply G in the vertical 
through (7, and combine it with P a at their in- 
tersection D. The resultant P should not cut 
the base RS in a point beyond the middle third 
(or, if this rule gives too massive a pier, take 
such a width that the pressure per square 
inch at S shall not exceed a safe value as 
computed from § 362.) After one or two 
trials a satisfactory width can be obtained. 
We should also be assured that the angle PD G is less 
than 30°. The horizontal joints above RS should also be 
tested as if each were, in turn, the lowest base, and if 
necessary may be inclined (like ran) to prevent slipping. 
On no joint should the maximum pressure per square inch 
be > than 1 / w the crushing strength of the cement. Abut- 
ments of firm natural rock are of course to be preferred 
where they can be had. If water penetrates under an 
abutment its buoyant effort lessens the weight of the lat- 
ter to a considerable extent. 




/ *G 



Fig. 



436 



MECHANICS OF ENGINEERING. 



362. Maximum Pressure Per Unit of Area When the Resultant 
Pressure Falls at Any Given Distance from the Middle ; according 
to Navier's theory of the distribution of the pressure ; see 
§ 346. Case I. Let the resultant pressure P, Fig. 400, (a), 




Fig. 400. 



Fig. 401. 



fall within the middle third, a distance = nd (< j4> d) 
from the middle of joint (d = depth of joint.) Then we 
have the following relations : 

p (the mean press, per. sq. m.),p m (max. press, persq. in.), 
and p n (least press, per sq. in.) are proportional to the lines 
h (mid. width), a (max. base), and c (min. base) respectively, 
of a trapezoid, Fig. 400, (5), through whose centre of gravity 
P acts. But (§ 26) 



d a—c • T/ 

nd=-^. i.e., n— 1 /^ 



a— i 



or a=h (6n-\-l) 



6 a-\-c h 

p m =p (6%+l). Hence the following table : 



If nd= yi d 

press. p m = 2 



%d 

5 A 



l Asd 

7. 



then the max. 

times the mean pressure. 



Case II. Let P fall outside the mid. third, a distance = 
nd (y }4> d) from the middle of joint. Here, since the 
joint is not considered capable of withstanding tension, 
we have a triangle, instead of a trapezoid. Fig. 401. First 
compute the mean press, per sq. in. 



■_ P (lbs.) 

P ~ (l-2») 18 d inches 
foot thick). 



or from this table : (lamina one 



ARCHES OF MASONRY. 



437 



For nd = 


^d 


&a 


6 J 


T\d 


-hd frd 


jP = 


1 P 

10 " d 


i p 

8 ' d 


1 P 

6 ' d 


1 P 

4 ' cl 


\ ' J, infinit ^ 



(d in inches and P in lbs.; with arch lamina 1 ft. in thickness.) 

Then the maximum pressure (at A, Fig. 401) p m , = 2p, 
becomes known, in lbs. per sq. in. 

362a. Arch-ring under Non-vertical Forces. — An example of 
this occurs when a vertical arch-ring is to support the pressure 
of a liquid on its extrados. Since water-pressures are always 
at right angles to the surface pressed on, these pressures on the 
extradosal surface of the arch-ring form a system of non-paral- 
lel forces which are normal to the curve of the extrados at 
their respective points of application and lie in parallel 
vertical planes, parallel to the faces of the lamina. We here 
assume that the extradosal surface is a cylinder (in the most 
general sense) whose rectilinear elements are "1 to the faces of 
the lamina. If, then, we divide the length of the extrados, 
from crown to each abutment, into from six to ten parts, the 
respective pressures on the corresponding surfaces are obtained 
by multiplying the area of each by the depth of its centre of 
gravity from the upper free surface of the liquid, and this 
product by the weight of a unit of volume of the liquid ; and 
each such pressure may be considered as acting through the 
centre of the area. Finally, if we find the resultant of each 
of these pressures and the weight of the corresponding portion 
of the arch-ring, these resultants form a series of non-vertical 
forces in a plane, for which an equilibrium polygon can then 
be passed through three assumed points by § 378a, these three 
points being taken in the crown-joint and the two abutment- 
joints. As to the " true linear arch" see § 359. 

As an extreme theoretic limit it is worth noting that if the 
extrados and intrados of the arch-ring are concentric circles; if 
the weights of the voussoirs are neglected ; and if the rise of 
the arch is very small compared with the depth of the crown 
below the water surface, then the circular centre-line of the 
arch-ring is the " true linear arch? 



438 MECHANICS OF ENGINEERING. 



CHAPTER XL 



ARCH-RIBS. 



364. Definitions and Assumptions. — An arch-rib (or elastic- 
arch, as distinguished from a block-work arch) is a rigid 
curved beam, either solid, or built up of pieces like a 
truss (and then called a braced arch) the stresses in which, 
under a given loading and with prescribed mode of sup- 
port it is here proposed to determine. The rib is sup- 
posed symmetrical about a vertical plane containing its 
axis or middle line, and the Moment of Inertia of any cross 
section is understood to be referred to a gravity axis of 
the section, which (the axis) is perpendicular to the said 
vertical plane. It is assumed that in its strained condi- 
tion under a load, the shape of the rib differs so little 
from its form when unstrained that the change in the ab- 
scissa or ordinate of any point in the rib axis (a curve) 
may be neglected when added (algebraically) to the co- 
ordinate itself ; also that the dimensions of a cross-section 
are small compared with the radius of curvature at any 
part of the curved axis, and with the span. 

385. Mode of Support. — Either extremity of the rib may be 
hinged to its pier (which gives freedom to the end-tangent- 
line to turn in the vertical plane of the rib when a load is 
applied); or may be fixed, i.e., so built-in, or bolted rigid- 
ly to the pier, that the end-tangent-line is incapable of 
changing its direction when a load is applied. A hinge 
may be inserted anywhere along the rib, and of course 



ARCH RIBS. 



439 



destroys the rigidity, or resistance to bending at that 
point. (A. hinge having its pin horizontal ~| to the axis of 
the rib is meant). Evidently no more than three such 
hinges could be introduced along an arch- rib between two 
piers ; unless it is to be a hanging structure, acting as a 
suspension-cable. 

366. Arch Rib as a Free Body. — In considering the whole 
rib free it is convenient, for graphical treatment, that no 
section be conceived made at its extremities, if fixed ; hence 
in dealing with that mode of support the end of the rib 
will be considered as having a rigid prolongation reach- 
ing to a point vertically above or below the pier junction, 
an unknown distance from it, and there acted on by a force 
of such unknown amount and direction as to preserve the 
actual extremity of the rib and its tangent line in the same 
position and direction as they really are. As an illustra- 
tion of this Eig. 402 
shows free an arch rib. 
ONB, with its extremi- 
ties and B fixed in the 
piers, with no hinges, q 
and bearing two 
loads P L and P 2 . The 
other jces of the sys- 
tem holding it in equi- fig. 402. 
librium are the horizontal and vertical components, of the 
pier reactions (H, V, H ni and V n ), and in this case of fixed 
ends each of these two reactions is a single force not in- 
tersecting the end of the rib, but cutting the vertical 
through the end in some point F (on the left ; and in G on 
the right) at some vertical distance c, (or d), from the end. 
Hence the utility of these imaginary prolongations OQF, 
and BRG, the pier being supposed removed. Compare 
Figs. 348 and 350. 

The imaginary points, or hinges, F and G, will be called 
abutments being such for the special equilibrium polygon 




440 



MECHAXICS OF EXGIXEEKLtfG. 



(dotted line), while and B are the real ends of the curved 
beam, or rib. 

In this system of forces there are five unknowns, viz.: V, 
V n9 H = H n9 and the distances c and d. Their determina- 
tion by analysis, even if the rib is a circular arc, is ex- 
tremely intricate and tedious ; but by graphical statics 
(Prof. Eddy's method ; see § 350 for reference), it is com- 
paratively simple and direct and applies to any shape of 
rib, and is sufficiently accurate for practical purposes. 
This method consists of constructions leading to the loca- 
tion of the " special equilibrium polygon " and its force 
diagram. In case the rib is hinged to the piers, the re- 
actions of the latter act through these hinges, Fig. 403, 
i.e., the abutments of the special 
equilibrium polygon coincide with 
the ends of the rib and B, and for 
a given rib and load the unknown 
quantities are only three V, V nf and 
H; (strictly there are four ; but IX 
= gives H n = H). The solution 
by analytics is possible only for ribs of simple algebraic 
curves and is long and cumbrous ; whereas Prof. Eddy's 
graphic method is comparatively brief and simple and is 
applicable to any shape of rib whatever. 

367. Utility of the Special Equilibrium Polygon and its force 
diagram. The use of locating these will now be illustrated 
[See § 332]. As proved in §§ 332 and 334 the compres- 
sion in each " rod " or segment of the " special equilibrium 
polygon" is the anti-stress resultant of the cross sections in 
the corresponding portion of the beam, rib, or other struc- 
ture, the value of this compression (in lbs. or tons) being 
measured by the length of the parallel ray in the force 
diagram. Suppose that in some way (to be explained sub- 
sequently) the special equilibrium polygon and its force 
diagram have been drawn for the arch -rib in Fig. 404 hav- 
ing fixed ends, and B, and no hinges ; required the elastic 
stresses in anv cross-section of the rib as at ra. Let the 




Fig. 403. 



AECH RIBS. 



441 




FiG. 404. 

ac&le of the force-diagram on the right be 200 lbs. to the 
inch, say, and that of the space -diagram (on the left) 30 ft. 
to the inch. 

The cross section m lies in a portion TK, of the rib, cor- 
responding to the rod or segment be of the equilibrium 
polygon; hence its anti-stress-resultant is a force B 2 acting 
in the line 6c, and of an amount given in the force-diagram. 
Now B 2 is the resultant of V, II, and P lt which with the 
elastic forces at m form a system in equilibrium, shown in 
¥ig, 405 ; the portion FO Tm being considered free. Hence 




Fig. 405. 



Fio. 406. 



taking the tangent line and the normal at m as axes we 
should have I (tang, comps.) = ; I (norm, comps.) = ; 
and I (moms, about gravity axis of the section at m) = 0, 
and could thus find the unknowns p u p 2 , and J, which ap- 
pear in the expressions ^.Pthe thrust, ^L the moment of 



442 



MECHANICS OF ENGINEERING. 



the stress-couple, and J the shear. These elastic stresses 
are classified as in § 295, which see. p i and p 2 are lbs. per 
square inch, «7is lbs., e is the distance from the horizontal 
gravity axis of the section to the outermost element of 
area, (where the compression or tension is p 2 lbs. per sq. 
in., as due to the stress-couple alone) while 1 is the " mo- 
ment of inertia " of the section about that gravity axis. 
[See §§ 247 and 295 ; also § 85]. Graphics, however, gives 
us a more direct method, as follows : Since R 2 , in the line 
be, is the equivalent of V, H, and P lf the stresses at m will 
be just the same as if R 2 acted directly upon a lateral pro- 
longation of the rib at T (to intersect bcFig. 405) as shown 
in Fig. 406, this prolongation Tb taking the place of TOF 
in Fig. 405. The force diagram is also reproduced here. 
Let a denote the length of the ~"\ from m's gravity axis 
upon be, and z the vertical intercept between m and be. 
For this imaginary free body, we have, 

from I (tang. compons.)=0, i? 2 cos a=p 1 F r 

and from J (norm. compons.)=0,i? 2 sin a=J 

while from - (moms, about) ) i t> Vol 

,-, ., • j? \ n f we have M^a = -^ . 

the gravity axis oi m) = 0, ) e 

But from the two similar triangles (shaded ; one of them 
is in force diagram) a : z : : H :B 2 .-. R 2 a—Hz, whence we 
may rewrite these relations as follows (with a general state- 
ment), viz.: 



If the Special Equilibrium Polygon and Its Force Diagram Have 
Been Drawn for a given arch-rib, of given mode of support, 
and under a given loading, then in any cross-section of the 
rib, we have (F = area of section): 

The projection of the proper 
ray(oi the force diagram) up- 
on the tangent line of the rib 
drawn at the given section. 



(L) The Thrust, Le., frF= 



ARCH RIBS. 443 



(2.) The Shear, i.e., J", = 
(upon which dependsthe 
shearing stress in the 
web). (See §§ 253 and 

256). 



The projection of the proper 

ray (of the force diagram) up- 
on the normal to the rib curve 
at the given section. 



<3.) The Moment of the 

stress couple, i.e.,^~ , = 
e 



The product (Hz) of the H 
(or pole-distance) of the force- 
diagram by the vertical dis- 
tance of the gravity axis of the 
section from the spec, equilib- 
h rium polygon. 
By the " proper ray " is meant that ray which is parallel 
to the segment (of the equil. polygon) immediately under 
or above which the given section is situated. Thus in 
Fig. 404, the proper ray for any section on TK is B 2 ; on 
KB, B 3 ; on TO, R Y . The projection of a ray upon any 
given tangent or normal, is easily found by drawing through 
each end of the ray a line "J to the tangent (or normal) ; 
the length between these ~\ 's on the tangent (or normal) is 
the force required (by the scale of the force diagram). We 
may thus construct a shear diagram, and a thrust diagram 
for a given case, while the successive vertical intercepts 
between the rib and special equilibrium polygon form a 
moment diagram. For example of the z of a point m is j4 
inch in a space diagram drawn to a scale of 20 feet to the 
inch, while H measures 2.1 inches in a force diagram con- 
structed on a scale of ten tons to the inch, we have, for the 
moment of the stress-couple at m, M=Hz= [2.1 X 10] tons 
x[ 1^x20] ft. =210 ft. tons. 

368. — It is thus seen how a location of the special equili- 
brium polygon, and the lines of the corresponding force- 
diagram, lead directly to a knowledge of the stresses in all 
the cross-sections of the curved beam under consideration, 
bearing a given load ; or, vice versa, leads to a statement 
of conditions to be satisfied by the dimensions of the rib, 
for proper security. 

It is here supposed that the rib has sufficient lateral 



444 



MECHANICS OF ENGINEERING. 



bracing (with others which lie parallel with it) to prevent 
buckling sideways in any part like a long column. Before 
proceeding to the complete graphical analysis of the differ- 
ent cases of arch -ribs, it will be necessary to devote the 
next few paragraphs to developing a few analytical rela- 
tions in the theory of flexure of a curved beam, and to 
giving some processes in " graphical arithmetic." 

369. Change in the Angle Between Two Consecutive Rib Tan- 
gents when the rib is loaded, as compared with its value 
before loading. Consider any small portion (of an arch 
rib) included between two consecutive cross-sections ; Eig. 
407. KHGW is its unstrained form. Let EA, = ds, be 
the original length of this portion of the rib axis. The 
length of all the fibres (|j to rib-axis) was originally =ds 
(nearly) and the two consecutive tangent-lines, at E and A, 
made an angle = dd originally, with each other. While 
under strain, however, all the fibres are shortened equally 
an amount dl l9 by the uniformly distributed tangential 
thrust, but are unequally shortened (or lengthened, accord- 
ing as they are on one side or the other of the gravity axis 
E, or A, of the section) by the system of forces making 
what we call the " stress couple," among which the stress 
at the distance e from the gravity axis A of the section is 
called p 2 per square inch ; so that the tangent line at A' 
now takes the direction AD "| to H' A G' instead of A f 
{we suppose the section at E to remain fixed, for conveni- 



h/h^f (Wt 



ML- 




Fig. 407. 



ARCH RIBS. 



445 



ence, since the change of angle between the two tangents 
depends on the stresses acting, and not on the new posi- 
tion in space, of this part of the rib), and hence the angle 
between the tangent-lines at E and A (originally = dd) is 
now increased by an amount CA'D = dip (or G' A'R — dtp); 
G'R' is the new position of GH. We obtain the value of 
dip as follows : That part (dl 2 ) of the shortening of the 
fibre at G, at distance e from A due to the force p 2 dF, is 

§ 201 eq. (1), dX 2 = ^r* But, geometrically, dk 2 also =ed<p, 



E 



Eed<p=p 2 ds 



(i.) 



But, letting M denote the moment of the stress-couple 
at section A (M depends on the loading, mode of support, 
etc., in any particular case) we know from § 295 eq. (6) that 

M=2l- 9 and hence by substitution in (1) we have 



d<p = 



Mds 
EI 



(2) 



[If the arch-rib in question has less than three hinges, 
the equal shortening of the fibres due to the thrust (of 
the block in last figure) p x F, will have an indirect effect on 
the angle dip. This will be considered later.] 

370. Total Change i.e. Cdip i n the Angle Between the End 

Tangents of a Rib, before and after loading. Take the ex- 
ample in Fig. 408 of a rib fixed at one end and hinged at 




Fig. 4 



446 



MECHANICS OF ENGINEERING. 



the other. When the rib is unstrained (as it is supposed 
to be, on the left, its own weight being neglected ; it is not 
supposed sprung into place, but is entirely without strain) 
then the angle between the end-tangents has some value 

d'= J dd= the sum of the successive small angles dd for 

each element ds of the rib curve (or axis). After loading, 
[on the right, Fig. 408], this angle has increased having 
now a value 



'+ f\, 



i.e., a value 



,+ /. 



B Mds 
EI 



(I.) 




Fig. 409. 



There must oe no hinge between 
and B. 

§ 371. Example of Equation (I) in Anal- 
ysis. — A straight, homogeneous, pris- 
matic beam, Fig. 409, its own weight 
neglected, is fixed obliquely in a wall. 
After placing a load P on the free end, 
required the angle between the end- 
tangents. This was zero before load- 
ing .*. its value after loading is 



:0 +,'=o+ -Jr/ia. 



By considering free a portion between and any ds of the 
beam, we find that M=Px=mom. of the stress couple. 
The flexure is so slight that the angle between any ds and 
its dx is still practically =a (§ 364), and .\ ds=dx sec a. 
Hence, by substitution in eq. (I.) we have 



«£,'=_! rMds = 
f EI Jo 



EI Jo 



P sec a r lc ^ a 



f 



: P ( cos ^ 2 [Compare with § 237], 
2EI L r 



AECH BIBS. 



447 



It is now apparent that if both ends of an arch rib are 
fixed, when unstrained, and the rib be then loaded (within 
elastic limit, and deformation slight) we must have 

B 

j {Mds-^-EI) — zero, since ^'=0. 

372. Projections of the Displacement of any Point of a Loaded 
Eib Relatively to Another Point and the Tangent Line at the Lat- 
ter. — (There must be no hinge between and B). Let 
be the point whose displacement is considered and B the 
other point. Fig. 410. If i?'s tangent-line is fixed while 
the extremity is not supported in any way (Fig. 410) 
then a load P put on, is displaced to a new position O a . 




Fig. 410. 



Fig. 411. 



Fig. 412. 



~With as an origin and OB as the axis of X, the projec- 
tion of the displacement OO n upon X, will be called dx f 
that upon Y, Ay. 

In the case in Fig. 410, O's displacement with respect to 
B and its tangent-line B T, is also its absolute displacement 
in space, since neither B nor B T has moved as the rib 
changes form under the load. In Fig. 411, however, the 
extremities and B are both hinged to piers, or supports, 
the dotted line showing its form when deformed under a 
load. The hinges are supposed immovable, the rib being 
free to turn about them without friction. The dotted line 
is the changed form under a load, and the absolute dis- 
placement of is zero ; but not so its displacement rela- 
tively to B and B's tangent BT, for BT has moved to a 
new position BT. To find this relative displacement con- 
ceive the new curve of the rib superposed on the old in 
a way that B and BT m&y coincide with their original po- 



448 



MECHANICS OF ENGINEERING. 



sitions, Fig. 412. It is now seen that O's displacement 
relatively to B and BT is not zero but — 00 u , and has a 
small Ax but a comparatively large Ay. In fact for this 
case of hinged ends, piers immovable, rib continuous be- 
tween them, and deformation slight, we shall write Ax — 
zero as compared with Ay, the axis X passing through OB), 

373. Values of the X and Y Projections of O's Displacement Rela- 
tively to B and B's Tangent ; the origin being taken at 0. 
Fig. 413. Let the co- 
ordinates of the dif- 
ferent points E, D, G, 
etc., of the rib, re- 
ferred to and an 
arbitrary X axis, be 
x and y, their radial 
distances from be- 
ing u (i.e., u for G, u* 
for D, etc.; in gener- 
al, u). OEDG is the a_.jl_ (s x ) 

unstrained form of the i * j 

rib, (e.g., the form it Fl ' G 413 

would assume if it lay flat on its side on a level platform, 
under no straining forces), while O n E"B'GB is its form 
under some loading, i.e., under strain. (The superposi- 
tion above mentioned (§ 372) is supposed already made if 
necessary, so that BT is tangent at B to both forms). 
Now conceive the rib OB to pass into its strained condi- 
tion by the successive bending of each ds in turn. The 
straining or bending of the first ds, BG, through the small 
angle dtp (dependent on the moment of the stress couple 
at G in the strained condition) causes the whole finite piece 
OG to turn about G as a centre through the same small 
angle dtp ; hence the point describes a small linear arc 
00* =dv, whose radius = u the hypothenuse of the x and 
y of 0, and whose value .*. is ov=udtp. 

Next let the section I), now at D*, turn through its- 
proper angle dtp* (dependent on its stress-couple) carrying 




AECH RIBS. 449 

with it the portion D'0\ into the position D 1 '0" \ making 
0' describe a linear arc 0' 0" =(3v)' =u'd(p', in which u' — 
the hjpothenuse on the x' and y' (of D), (the deformation 
is so slight that the co-ordinates of the different points 
referred to and X are not appreciably affected). Thus, 
each section having been allowed to turn through the an- 
gle proper to it, finally reaches its position, U , of dis- 
placement. Each successive dv, or linear arc described by 
0, has a shorter radius. Let dx, (dx)', etc., represent the 
projections of the successive (ov)'s upon the axis X; and 
similarly dy, (dy)' etc., upon the axis Y. Then the total X 
projection of the curved line . . . . Q will be 

Ax— J dx and similarly Ay = joy . . . (1) 

But o v = u d <p, and from similar right-triangles, 
d x ; dv : : y : u and by : dv : : x : u .*. dx = ydcp and oy=xdcp ; 
whence, (see (1) and (2) of §369) 

A X = f Sx = fyd v =£JMi . . . (II.) 

& ndAy=fdy=fxd<p = £^- . . . (HI.) 

If the rib is homogeneous E is constant, and if it is of 
constant cross-section, all sections being similarly cut by 
the vertical plane of the rib's axis (i.e., if it is a " curved 
prism ") /, the moment of inertia is also constant. 

374. Recapitulation of Analytical Relations, for reference* 

(Not applicable if there is a hinge between and B) 

Total Change in Angle between ) _ /^Mds 

tangent-lines and B ) ~~J ~EI • ° • • (• L ) 

The X-Projection of O's Displacement "1 

Relatively to B.and B's tangent- .^ 

line ; {the origin being at 0) I = I * ■ • • • (■«■.; 

and the axes X and 7 H to j Jo El 

each other) I 



450 



MECHANICS OF ENGINEERING. 



The Y-Projection of 0's Displacement, ) 
etc., as above. 



}=X 



*31xds 



(in.) 



Here x and y are the co-ordinates of points in the rib- 
curve, ds an element of that curve, M the moment of the 
stress-couple in the corresponding section as induced by 
the loading, or constraint, of the rib. 

(The results already derived for deflections, slopes, etc., 
for straight beams, could also be obtained from these 
formulae, I., II. and III. In these formulae also it must 
be remembered that no account has been taken of the 
shortening of the rib-axis by the thrust, nor of the effect 
of a change of temperature.) 



374a. Resumfe of the Properties of Equilibrium Polygons and 
their Force Diagrams, for Systems of Vertical Loads. — See §§ 335 
to 343. Given a system of loads or vertical forces, P l3 P 2 , 

etc., Fig. 414, and 
two abutment verti- 
cals, F' and G' ; if 
we lay off, vertically, 
to form a " load- 
line," 1 . . 2 = JPi, 2. . . 
S=P 2 , etc., select any 
Pole, Q lf and join X 
... 1, Ox ... 2, etc. ; 
also, beginning at 
Fi g- 4i4. an y point F x in the 

vertical F\ if we draw F l . . . a || to X . . 1 to intersect the 
line of Pi ; then ab || to Y . . 2, and so on until finally a 
point 6rj, in G', is determined ; then the figure F l .dbc G x is 
an equilibrium polygon for the given loads and load verti- 
cals, and Oi . . . 1234 is its " force diagram." The former 
is so called because the short segments F x a ab, etc., if 
considered to be rigid and imponderable rods, in a vertical 
plane, hinged to each other and the terminal ones to abut- 
ments F x and G u would be in equilibrium under the given 
loads hung at the joints. An infinite number of equilib- 




ARCH-RIBS. 451 

rium polygons may be drawn for the given loads and 
abutment-verticals, by choosing different poles in the force 
diagram. [One other is shown in the figure ; 2 is its 
pole. {F l Gi and F 2 G 2 are abutment lines.)] For all of 
these the following statements are true : 

(1.) A line through the pole, 11 to the abutment line cuts 
the load-line in the same point n', whichever equilibrium 
polygon be used ( /. anyone will serve to determine n f ). 

(2.) If a vertical CD be drawn, giving an intercept z' in 
each of the equilibrium polygons, the product Hz' is the 
same for all the equilibrium polygons. That is, (see Fig. 
414) for any two of the polygons we have 

H X :H 2 :: z/ : */ ; or H 2 z 2 ' = H, z x '. 

(3.) The compression in each rod is given by thai 
" ray " (in the force diagram) to which it is parallel. 

(4.) The " pole distance " H t or ~| let fall from the pole 
upon the load-line, divides it into two parts which are the 
vertical components of the compressions in the abutment- 
rods respectively ( the other component being horizontal) ; 
H is the horizontal component of each (and, in fact, of 
each of the compressions in all the other rods). The 
compressions in the extreme rods may also be called the 
abutment reactions (oblique) and are given by the extreme 
rays. 

(5.) Three Points [not all in the same segment (or rod)] 
determine an equilibrium polygon for given loads. Hav- 
ing given, then, three points, we may draw the equilibrium 
polygon by §341. 

375. Summation of Products. Before proceeding to treat 
graphically any case of arch-ribs, a few processes in 
graphical arithmetic, as it may be called, must be pre- 
sented, and thus established for future use. 

To make a summation of products of two factors in each 
by means of an equilibrium polygon. 



452 



MECHANICS OF ENGINEERING. 



Construction. Suppose it required to make the summa- 
tion I (x z) i. c, to sum the series 



x l z 1 +x 2 z 2 +x 3 z 5 + 



by graphics. 



Having first arranged the terms in the order of magni- 
tude of the x's, we proceed as follows : Supposing, for 
illustration, that two of the z'& (z B and z 4 ) are negative 
(dotted in figure) see Fig. 415. These quantities x and z 
may be of any nature whatever, anything capable of being 
represented by a length, laid off to scale. 

First, in Fig. 
416, lay off the 
a's in their 
order, end to 
end, on a ver- 
tical load-line 
taking care to 
lay off % and 
z± upivard in 
their turn. 
Take any con- 
venient pole 

having pre- 




... '-Z^^ L 



Fig. 415. 

; draw the rays . 



Fig. 416. 

1, ... 2, etc.; then, 
viously drawn vertical lines whose horizontal distances 
from an extreme left-hand vertical F f are made = x Li x, t 
a? 3 , etc., respectively, we begin at any point F, in the verti- 
cal F\ and draw a line || to ... 1 to intersect the x l ver- 
tical in some point ; then l 7 2' II to . . . 2, and so on, fol- 
lowing carefully the proper order. Produce the last seg- 
ment (6' ... G in this case) to intersect the vertical F' in 
some point K. Let KF =h (measured on the same scale 
as the as's), then the summation required is 

2f (xz) = m. 

H is measured on the scale of the z's, which need not be 
the same as that of the x's ; in fact the g's may not be the 
same kind of quantity as the as's. 

[Proof. — From similar triangles 3: z L :: x v : h lf .*. x ] z l = Hk^ ; 
and " " " H\ z 9 :: x 2 : k 2i .'. x 2 z 2 =Bk 2 . 






AUCH-KIBS. 



453 



and so on. But H (k l ~{-k 2 -j-etc.)=HxFK=Hk]. 

376. Gravity Vertical. — From the same construction in 
Pig. 415 we can determine the line of action (or gravity 
vertical) of the resultant of the parallel vertical forces z lt 
«2, etc. (or loads); by prolonging the first and last segments 

to their intersection at 
(j. The resultant of the 
system of forces or loads 
acts through C and is 
vertical in this case ; its 
value being == I (2), 
that is, it = the length 
1 ... 7 in the force dia- 
gram, interpreted by the 
proper scale. It is now 
supposed that the s's 
represent forces, the x's 
being their respective 
lever arms about F. If 
the ?.'s represent the 
areas of small finite por- 
tions of a large plane 
figure, we may find a 
gravity-line (through C) 
of that figure by the 
above construction; each 
z being-applied through 
the centre of gravity of 
its own portion. 

Calling the distance 
x between the verticals 
through C and F } we 
have also x . I (z) = 
I (xz) because I (z) is 
the resultant of the || z's. 
ki This is also evident from 
the proportion (similar 
triangles) 
H : (1 . . 7) :: x : k 




454 MECHANICS OP ENGINEERING. 

376 a. Moment of Inertia (of Plane Figure) by Graphics.— Fig. 
416 a. I N = ? First, for the portion on right. Divide OB 
into eqnal parts each = Ax. Let z lf z 2 , etc., be the middle 
ordinates of the strips thus obtained, and x lt etc. their 
abscissas (of middle points). 

Then we have approximately 

J N for OB — Ax.z^x^ Ax.z 2 x 2 2 -\- 

=Jx[(z 1 x 1 )x 1 -f(z 2 x 2 )x 2 + ...].. (1) 

But by §375 we may construct the products z l x l ,z 2 x 2 , etc., 
taking a convenient IF , (see Fig. 416, (6)), and obtain k l9 k 2 , 
etc., such that z x x x = H'ki, z 2 x 2 = H'k 2 , etc. Hence eq. (1) 
becomes : 

lj for OB SL-p-pTox.=H f Ax[k 1 x 1 -\-k 2 x 2 -\- ...]... (2) 

By a second use of § 375 (see Fig. 416 c) we construct Z,„ 
such that \x x -\- k 2 x 2 +.... = H"l [R" taken at con- 
venience]. .*. from eq. (2) we have finally, (approx.), 

i" N for OB=B'H"lAx (3) 

For example if OB = 4 in., with four strips, Ax would = 
1 in.; and if IF = 2 in., B" = 2 in., and I = 5.2 in., then 

7 N for OB = 2x2x5.2x1.0=20.8 biquad. inches. 

The J N for OL, on the left of N, is found in a similar 
manner and added to I N for OB to obtain the total 7 N . The 
position of a gravity axis is easily found by cutting the 
shape out of sheet metal and balancing on a knife edge ; or 
may be obtained graphically by § 336 ; or 376. 

377. Construction for locating a line vm (Fig. 417) at (a), in 
the polygon FG in such a position as to satisfy the two 
following conditions with reference to the vertical inter- 
cepts at 1, 2, 3, 4, and 5, between it and the given points 
1, 2, 3, etc., of the perimeter of the polygon. 



ARCH-RIBS. 



455 



Condition I. — (Calling these intercepts u lt u 2 , etc., and their 
horizontal distance-s from a given vertical F, x lf x. 2i etc.) 

I (u) is to = ; i.e., the sum of the positive ^'s must be 
numerically = that of the negative (which here are at 1 
and 5). An infinite number of positions of vm will satisfy 
condition I. 



Condition II. — 2 (ux) is to = ; i.e., the sum of the 

jy ! 1 r-—^m nioments of the positive us 

about F must = that of the 

G ( a ) negative w's. i.e., the moment 
of the resultant of the posi- 

m tive w's must = that of the 
resultant of the negative ; 

G and .*. (Condit. I being 

m already satisfied) these two 
resultants must be directly 

g( c ) opposed and equal. But the 
ordinates u in (a) are indi- 

m vidually equal to the differ- 
ence of the full and dotted 
ordinates in (b) with the 
same cc's .*. the conditions 
may be rewritten : 

I. 2 (full ords. in (b)) = 
I (dotted ords. in (&)) 

II. I [each full ord. in (b) 
X its x~\ = I [each dotted 
ord. in (b) x its x] i.e., the 

Fig. 417. centres of gravity of the full 

and of the dotted in (b) must lie in the same vertical. 

Again, by joining vG, we may divide the dotted ordi- 
nates of (b) into two sets which are dotted, and broken, re- 
spectively, in (c) Then, finally, drawing in (d), 

R, the resultant of full ords. of (c) 
T, " " " broken ' 

T', " " " dotted ' 




(d) 



(C 4( 
t< (6 



456 MECHANICS OF ENGINEERING. 

we are prepared to state in still another and final form the 
conditions which vm must fulfil, viz. : 

(I.) T-\-T' must = R; and (II.) The resultant of T 
and T' must act in the same vertical as R. 

In short, the quantities T, T', and R must form a bal- 
anced system, considered as forces. All of which amounts 
practically to this : that if the verticals in which T and T' 
act are known and R be conceived as a load supported by 
a horizontal beam (see foot of Fig. 417, last figure) resting 
on piers in those verticals, then T and T' are the respec- 
tive reactions of those piers. It will now be shown that the 
verticals of T and T' are easily found, being independent of 
the position of vm; and that both the vertical and the mag- 
nitude of R, being likewise independent of vm, are deter- 
mined with facility in advance. For, if v be shifted up 
or down, all the broken ordinates in (c) or (d) will change 
in the same proportion (viz. as vF changes), while the 
dotted ordinates, though shifted along their verticals, do 
not change in value ; hence the shifting of v affects neither 
the vertical nor the value of T' , nor the vertical of T. 
The value of T, however, is proportional to vF. Similar- 
ly, if m be shifted, up or down, T' will vary proportionally 
to mG, but its vertical, or line of action, remains the same. 
T is unaffected in any way by the shifting of m. R, de- 
pending for its value and position on the full ordinates of 
(c) Fig. 417, is independent of the location of vm. We 
may .•. proceed as follows : 

1st. Determine R graphically, in amount and position, 
by means of § 376. 

2ndly. Determine the verticals of T and T f by any trial 
position of vm (call it v 2 m 2 ), and the corresponding trial 
values of T and T' (call them T 2 and T' 2 ). 

3rdly. By the fiction of the horizontal beam, construct 
(§ 329) or compute the true values of T and T ', and then 
determine the true distances vF and mG by the propor- 
tions 

vF : vJF : : T : T 2 and mG : m 2 G : : T : V* 



ARCH-RIBS. 



457 



Example of this. Fig. 418. (See Fig. 417 for s and L) 

From A toward B in (e) Fig. 418, lay off the lengths (or 
lines proportional 
to them) of the full 
ordinates 1, 2, etc., 
of (/). Take any 
pole 0i, and draw the 
equilibrium poly- 
gon (f)' and pro- 
long its extreme seg- 
ments to find G and 
thus determine R's 
vertical. R is repre- 
sented by AB. In 
■(g) [same as (/) but 
shifted to avoid 
complexity of lines] 
draw a trial v 2 m 2 and 
join v 2 G 2 . Deter- 
mine the sum T 2 of j *S] Er 
the broken ordi- f fig. 418. 
nates (between v 2 G 2 ana F 2 G 2 ) and its vertical line of ap- 
plication, precisely as in dealing with R ; also T\ that of 
the dotted ordinates (five) and its vertical. Now the true 
T=Rtj-(s+t) and the true T'=Rs+(s+t). Hen ce com- 
pute vF=(T-^T 2 )v^F 2 and ^G=(T'+T\) m 2 G 2 , and by 
laying them off vertically upward from F and G respec- 
tively we determine v and m, i.e., the line vm to fulfil the 
conditions imposed at the beginning of this article, rela- 
ting to the vertical ordinates intercepted between vm and 
given points on the perimeter of a polygon or curve. 

Note (a\ If the verticals in which the intercepts lie are 
equidistant and quite numerous, then the lines of action 
of T 2 and T' 2 will divide the horizontal distance between 
F and G into three equal parts. This will be exactly true 
in the application of this construction to § 390. 

Note (b). Also, if the verticals are symmetrically placed 
about a vertical line, (as will usually be the case) v 2 m 2 is 




458 



MECHANICS OF ENGINEERING. 



best drawn parallel to FG, for then T, and T\ will be 
equal and equi-distant from said vertical line. 

378. Classification of Arch-Ribs, or Elastic Arches, according 
to continuity and modes of support. In the accompany- 
ing figures the full curves show the unstrained form of the 
rib (before any load, even its own weight, is permitted to 
come upon it) ;the dotted curve shows its shape (much ex- 
aggerated) when bearing a load. For a given loading 
Three Conditions must be given to determine the special 
equilibrium polygon (§§ 366 and 367). 

Class A. — Continuous rib, free to slip laterally on the 
piers, which have smooth horizontal surfaces, Fig. 420. 

This is chiefly of theoretic interest, its consideration 
being therefore omitted. The pier reactions are neces- 
sarily vertical, just as if it were a straight horizontal 
beam. 

Class B. Rib of Three Hinges, two at the piers and one 
intermediate (usually at the crown) Fig. 421. Fig. 36 also 
is an example of this. That is, the rib is discontinuous 
and of two segments. Since at each hinge the moment of 
the stress couple must be be zero, the special equilibrium 
polygon must pass through the hinges. Hence as three 
points fully determine an equilibrium polygon for given 
load, the special equilibrium is drawn by § 341. 




Fig. 420. 



Fig. 421. 



[§ 378a will contain a construction for arch-ribs of three 
hinges, when the forces are not all vertical.] 

Class C. Rib of Two Hinges, these being at the piers, the 
rib continuous between. The piers are considered im- 
movable, i.e., the span cannot change as a consequence of 
loading. It is also considered that the rib is fitted to its 



ARCH EIBS. 



459 



hinges at a definite temperature, and is then under no con- 
straint from the piers (as if it lay flat on the ground), not 
even its own weight being permitted to act when it is fi- 
nally put into position. When the " false works " 
or temporary supports are removed, stresses are in- 
duced in the rib both by its loading, including its 
own weight, and by a change of temperature. Stresses 
due to temperature may be ascertained separately and 
then combined with those due to the loading. [Classes 
A and B are not subject to temperature stresses.] Fig. 

422 shows a rib of two hinges, 
at ends. Conceive the dotted 
curve (form and position un- 
der strain) to be superposed 
on the continuous curve 
(form before strain) in such 
a way that B and its tangent 
line (which has been dis- 
placed from its original position) may occupy their pre- 
vious position. This gives us the broken curve n B. 00 Q 
is .*. O's displacement relatively to B and B's tangent. 
Now the piers being immovable n B (right line) =0B ; i.e., 
the X projection (or Ax) of 00 n upon OB (taken as an axis 
of X) is zero compared with its Ay. Hence as one condi- 
tion to fix the special equilibrium polygon for a given load- 
ing we have (from § 373) 




Fig. 422. 



f\Mydi 



.ET]=0 



(1) 



The other two are that the [ must pass through . (2) 
special equilibrium polygon j " " " B . (3) 

Class D. Rib with Fixed Ends and no hinges, i.e., continu- 
ous. Piers immovable. The ends may be fixed by being 
inserted, or built, in the masonry, or by being fastened to 
large plates which are bolted to the piers. [The St. Louis 
Bridge and that at Coblenz over the Rhine are of this 
class.] Fig. 423. In this class there being no hinges we 



460 



MECHANICS OF ENGLNEEKING. 



have no point given in advance through which the special 
equilibrium polygon must pass. However, since O's dis- 
placement relatively (and absolutely) to B and B's tangent 
is zero, both Ax and Ay [see § 373] — zero. Also the tan- 
gent-lines both at and B being 
fixed in direction, the angle be- 
tween them is the same under 
loading, or change of temperature, 
as when the rib was first placed 
in position under no strain and at 
a definite temperature. 
Hence the conditions for locating the special equilibrium 
polygon are 




Fig. 423. 



/. 



Mds _ q . p Myds _ q 



x 



'Mxds 



0. 



EI Jo El Jo EI 

In the figure the imaginary rigid prolongations at the 
ends are shown [see § 366]. 

Other designs than those mentioned are practicable 
(such as : one end fixed, the other hinged ; both ends fixed 
and one hinge between, etc.), but are of unusual occur- 
rence. 

378a. Rib of Three Hinges. Forces not all Vertical. If the 
given rib of three hinges upholds a roof, the wind-press- 
ure on which is to be considered as well as the weights of 
the materials composing the roof-covering, the forces will 
not all be vertical. To draw the special equil. polygon in 

/ such a case the following 

/ R -» construction holds : Re- 

quired to draw an equilib- 
rium polygon, for any 
plane system of forces, 
through three arbitrary 
\ points, A, p and B ; Fig. 
~B423a. Find the line of 
action of i? 1? the resultant 
of all the forces occurring 
between A and p; also, 




\ / 



Fig. 423a. 



ARCH-KIBS. 461 

that of R 2 , the resultant of all forces between ,p and B ; 
also the line of action of R, the resultant of R Y and R 2 , [see 
§ 328.] Join any point M in R with A and also with B, 
and join the intersections JVand 0. Then A N will be the 
direction of the first segment, B that of the last, and 
NO itself is the segment corresponding to p (in the de- 
sired polygon) of an equilibrium polygon for the given 
forces. See § 328. If A N' p 0' B are the corresponding 
segments (as yet unknown) of the desired equil. polygon, 
we note that the two triangles MNO and M'N'O', having 
their vertices on three lines which meet in a point [i.e., R 
meets R Y and R 2 in (7'], are homological [see Prop. VII. of 
Introduc. to Modern Geometry, in Chauvenet's Geometry,] 
and that . ■ . the three intersections of their corresponding 
sides must lie on the same straight line. Of those inter- 
sections we already have A and B, while the third must be 
at (7, found at the intersection of AB and NO. Hence by 
connecting C and p, we determine N and 0' . Joining 
N'A and O'B, the first ray of the required force diagram will 
be || to NA, while the last ray will be || to O'B, and thus 
the pole of that diagram can easily be found and the cor- 
responding equilibrium polygon, beginning at A, will pass 
through p and B. 
(This general case includes those of §§ 341 and 342.) 

379. Arch-Rib of two Hinges; by Prof. Eddy's Method.* 
[It is understood that the hinges are at the ends.] Ke- 
quired the location of the special equilibrium polygon. We 
here suppose the rib homogeneous (i.e., the modulus of 
Elasticity E is the same throughout), that it is a " curved 
prism " (i.e., that the moment of inertia i" of the cross- 
section is constant), that the piers are on a level, and that 
the rib-curve is symmetrical about a vertical line. Fig. 

424. For each point m of the rib 
curve we have an x and y (both 
known, being the co-ordinates of 
the point), and also a z (intercept 
-t between rib and special equilib- 
rium polygon) and a z f (intercept 

* P. 25 of Prof. Eddy's book ; see reference in preface of this work. 




462 



MECHANICS OF ENGINEERING. 



between the spec. eq. pol. and the axis X (which is OB). 
The first condition given in § 378 for Class C may be 
transformed as follows, remembering [§ 367 eq. (3)] that 
M == Hz at any point ra of the rib (and that EI is con- 
stant). 

Ij. £ My da = 0, i.e., ^ j\yds = . • . j\yds = 



but 
z=y - z' 



• C (y~ z')yds=0;Le., f yyds = f yz'ds . (1) 

t/o do do 



In practical graphics we can not deal with infinitesimals ; 
hence we must substitute As a small finite portion of the 
rib-curve for ds; eq. (1) now reads 2* yy As = 2* B yz' As. 
But if we take all the As's equal, As is a common factor 
and cancels out, leaving as a final form for eq. (1) 

z \ yy y=z*{y S ') . . . ay 

The other two conditions are that the special equilibrium 
polygon begins at and ends at B. (The subdivision of 
the rib-curve into an even number of equal As's will be ob- 
served in all problems henceforth.) 

379a. Detail of the Construction. Given the arch -rib B, 
Pig. 425, with specified loading. Divide the curve into 



1 2 



i f 

4 5 



ll I 



X 



-f 



7 8 | 



II 



\ h i 



\ 



/ f 3 4 ?5 \ ' 





Fig. 425. 



ARCH RIBS. 



463 



eight equal ^s's and draw a vertical through the middle 
of each. Let the loads borne by the respective As's be 
P lt P 2 , etc., and with them form a vertical load-line A C to 
some convenient scale. With any convenient pole 0" 
draw a trial force diagram 0" AC, and a corresponding 
trial equilibrium polygon F G, beginning at any point in 
the vertical F. Its ordinates «/', %"> etc., are propor- 
tional to those of the special equil. pol. sought (whose 
abutment line is OB) [§ 374a (2)]. We next nse it to de- 
termine n r [see § 374a]. We know that OB is the " abut- 
ment-line " of the required special polygon, and that . * . 
its pole must lie on a horizontal through n' . It remains 
to determine its H, or pole distance, by equation (1/ just 
given, viz. : 2\ 8 yy — I}yz f . First by § 375 find the value 
of the summation If(yy), which, from symmetry, we may 
write = 2I 1 \yy)=2[y l y l -\-y 2 y 2 -{-y 3 y 3 +y i y i ] 




Hence, Fig. 426, we obtain 

1\ (yy)=2 [R k] 



Next, also by § 375, see Fig. 
427, using the same pole dis- 
tance H as in Fig. 426, we 
find 



i.e. 






2/1*1 +2/2*2 

H n h". 



Again, since If (yz")= y 8 z 8 " 
+ 2/7*7" + 2/6*6" + 2/5*5" which 
from symmetry (of rib) 

= 2/l*8" + 2/2*/'+2/3*6" + 2/4*5'', 



we obtain, Fig. 428, 
21 (2/*") = BX\ (same 
and .*. 



#0); 



2\ (yz")=H (k x "+K"). If now we find that V+ V=2fc, 



464 MECHANICS OF ENGINEERING. 

the condition 2\ (yy) = 2f (yz") is satisfied, and the pole 
distance of our trial polygon in Fig. 425, is also that of 
the special polygon sought; i.e., the z" 's.are identical in 
value with the s"s of Fig. 424. In general, of course, we 
do not find that &/'+&/' = 2k. Hence the z" 's must all 
be increased in the ratio 2k: (V+V) to become equal to 
the s"s. That is, the pole distance H of the spec, equih 
polygon must be 

tt_ &/'+ V rrff (in which H" — the pole distance of the 
2k trial polygon) since from §339 the ordi- 

nates of two equilibrium polygons (for the same loads) 
are inversely as their pole distances. Having thus found 
the If oi the special polygon, knowing that the pole must 
lie on the horizontal through n' 9 Fig. 425, it is easily 
drawn, beginning at 0. As a check, it should pass through 
B. 

For its utility see § 367, but it is to be remembered that 
the stresses as thus found in the different parts of the 
rib under a given loading, must afterwards be combined 
with those resulting from change of temperature and the 
shortening of the rib axis due to the tangential thrusts, 
before the actual stress can be declared in any part. 

[Note. — If the " moment of inertia," 7, of the rib-sec- 
tion is different at different sections, i.e., if I is variable, 

foreq. (1)' we may write 2%((yM-)=IH(y. Z ') . . . (1)' 

(where n = -, 1 being the moment of inertia of a particu- 
lar section taken as a standard and I that at any section 

of rib) and in Fig. 426, use the U- of each As instead of y 

n 

in the vertical " load-line," and — for z" in Figs. 427 and 

n 

428]. 



AHCH-MB3. 



465 



380. Arch Rib of Fixed Ends and no Hinges,— Example of 
Class D. Prof. Eddy's Method.* As before, E and I are 
constant along the rib Piers immovable. Rib curve 
symmetrical about a vertical line. Fig. 429 shows such a 
rib under any loading. Its span is OB, which is taken as 
an axis X. The co-ordinates of any point mf of the rib 
curve are x and y, and z is the vertical intercept between 
m' and the special equilibrium polygon (as yet unknown, 
but to be constructed). Prof. Eddy's method will now be 

given for finding tha spe- 
cial equil. polygon. The 
fchree conditions it 
must satisfy (see § 378, 
Class D, remembering 
that E and /are constant 
and that M = Ih isjxa 
§ 367) are 




h^3) 



Fig. 429. 



Czds= ; Cxzds= ; and Cyzds—0 

e/o e/o e/o 



(i) 



Now suppose the auxiliary reference line (straight) vm 
to have been drawn satisfying the requirements, with 
respect to the rib curve that 



Pz'ds = ; and Cxz'ds = 

e/o Jo 



(2) 



in which z' is the vertical distance of any point rrv from 
vm and x the abscissa of m' from 0. 

From Fig. 429, letting z" denote the vertical intercept 
(corresponding to any m') between the spec, polygon and 
the auxiliary line vm, we have z—z'—z", hence the three 
conditions in (1.) become 

j(z'-z")ds=0; i.e., see eqs. (2) C* z"ds=0 . . . (3) 



* P. 14 of Prof Eddv's book ; see reference in preface of this work. 



466 MECHANICS OF ENGINEERING. 

/ x (z T —z t ')ds=0 ; i.e., see eqs. (2) / xz"ds=o (4j 

do do 

•rfjfr-O^OA^^^A-yJtf* . (5) 

provided vm has been located as prescribed. 

For graphical purposes, having subdivided the rib curve 
into an even number of small equal Js's, and drawn a verti- 
cal through the middle of each, we first, by § 377, locate 
vm to satisfy the conditions 

27(»')=0 and 2?(a» / )=0 . . (6) 

(see eq. (2) ; the As cancels out) ; and then locate the 
special equilibrium polygon, with vm as a reference-line, 
by making it satisfy the conditions. 

2?(*")=0 • (7); IK)=0 • (8); Z*(yz") = I*(yz') . (9) 

(obtained from eqs. (3), (4), (5) by putting ds=As, and can- 
celling). 

Conditions (7) and (8) may be satisfied by an infinite 
number of polygons drawn to the given loading. Any one 
of these being drawn, as a trial polygon, we determine for it 
the value of the sum l^yz") by § 375, and compare it with 
the value of the sum I*(yz') which is independent of the 
special polygon and is obtained by § 375. [N.B. It must 
be understood that the quantities (lengths) x, y, z, z', and z' f , 
here dealt with are those pertaining to the verticals drawn 
through the middles of the respective ^s's, which must be 
sufficiently numerous to obtain a close result, and not to 
the verticals in which the loads act, necessarily, since these 
latter may be few or many according to circumstances, see 
Fig. 429]. If these sums are not equal, the pole distance 
of the trial equil. polygon must be altered in the proper 
ratio (and thus change the z'"s in the inverse ratio) neces- 
sary to make these sums equal and thus satisfy condition 
(9). The alteration of the s'"s, all in the same ratio, will 



ARCH-RIBS. 



467 



not interfere with conditions (7) and (8) which are already 
satisfied. 

381. Detail of Construction of Last Problem. Symmetrical Arch- 
Rib of Fixed Ends. — As an example take a span of the St. 
Louis Bridge (assuming /constant) with " live load" cov- 
ering the half span on the left, Eig. 430, where the vertical 




Bd-3=sM=y 



Fig. 430. 



scale is much exaggerated for the sake of distinctness*. 
Divide into eight equal Js's. (In an actual example sixteen 
or twenty should be taken.) Draw a vertical through the 



* Each arch-rib of the St. Louis bridge is a built up or trussed rib of steel about 52ft 
ft. span and 52 ft rise, in the form of a segment of a circle . Its moment of inertia, 
however, is not strictly constant, the portions near each pier, of a length equal to one 
twelfth of the span, having a value of / one-half greater than that of the remainder o£ 
the arc. 



468 MECHANICS OF ENGINEERING. 

middle of each As. P 1} etc., are the loads coming upon the 
respective Js's. 

First, to locate vm, by eq. (6) ; from symmetry it must 
be horizontal. Draw a trial vm (not shown in the figure) 
and if the (+s')'s exceed the ( — z')'s by an amount z Q \ the 

true vm will lie a height - zj above the trial vm (or be- 

n K 

low, if vice versa) ; n~ the number of Js's. 

Now lay off the load-line on the right, (to scale), 
take any convenient trial pole 0'" and draw a correspond- 
ing trial equil. polygon F'"G"\ In F'"G f ", by § 377, 
locate a straight line v"m" so as to make l*(z"')=0 and 
l*(xz"')=0 (see Note (b) of § 377). 

[We might now redraw F'"G'" in such a way as to bring 
v'"m'" into a horizontal position, thus : first determine a 
point n'" on the load-line by drawing 0"'n'" || to v'"m'" take 
a new pole on a horizontal through ri", with the same 
H"', and draw a corresponding equil. polygon ; in the lat- 
ter v' f 'm"' would be horizontal. We might also shift this 
new trial polygon upward so as to make v"m''' and vm 
coincide. It would satisfy conditions (7) and (8), having 
the same s""s as the first trial polygon ; but to satisfy con- 
dition (9) it must have its z""s altered in a certain ratio, 
which we must now find. But we can deal with the indi- 
vidual z" n s just as well in their present positions in Fig. 
430.] The points E and L in vm, vertically over E f " and 
L"' in v'"m'", are now fixed ; they are the intersections of the 
special polygon required, with vm. 

The ordinates between v'"m'" and the trial equilibrium 
polygon have been called z'" instead of z"\ they are pro- 
portional to the respective s"'s of the required special 
polygon. 

The next step is to find in what ratio the (gf") 9 & need to 
be altered (or H'" altered in inverse ratio) in order to be- 
come the (s")' s > i* e *> i n or der to fulfil condition (9), viz.: 



ARCH-BIBS. 



469 




Z\{yz")=Z\(yz>) . (9) 

This may be done pre- 
cisely as for the rib with 
two hinges, but the nega- 
tive (zf'ys must be prop- 
erly considered (§ 375) 
See Fig. 431 for the de- 
tail. Negative 3"s or s""s 
point upward. 

From Fig. 431a 

Il(yz')=H k 
»*. from symmetry 

l%yz')=2H k 
From Fig. 4315 we have 



Pig. 431. 

and from Fig. 431c 



[The same pole distance H Q is taken in all these construc- 
tions] .-. I\{yz")=Hlk+k v ). 

If, then, H (k x -{-k v ) = 2H k condition (9) is satisfied by the 
e""s. If not, the true pole distance for the special equil. 
polygon of Fig. 430 will be 



H= 



^i~r^f 



2k 



H' 



With this pole distance and a pole in the horizontal through 
to,"' (Fig. 430) the force diagram may be completed for the 
required special polygon ; and this latter may be con- 
structed as follows : Beginning at the point E, in vm f 
through it draw a segment || to the proper ray of the force 
diagram. In our present figure (430) this " proper ray " 
would be the ray joining the pole with the point of meet- 
ing of P 2 and P 3 on the load-line. Having this one seg- 



470 MECHANICS OF ENGINEERING. 

ment of the special polygon the others are added in an 
obvious manner, and thus the whole polygon completed. 
It should pass, through L, but not and B. 

For another loading a different special equil. polygon 
would result, and in each case we may obtain the thrust, 
shear, and moment of stress couple for any cross-section of 
the rib, by § 367. To the stresses computed from these, 
should be added (algebraically) those occasioned by change 
of temperature and by shortening of the rib as occasioned 
by the thrusts along the rib. These " temperature 
stresses," and stresses due to rib-shortening, will be con- 
sidered in a subsequent paragraph. They have no exist- 
ence for an arch-rib of three hinges. 

[Note. — If the moment of inertia of the rib is variable 

z' z" 

we put — for z' and — for z" in equation (6), (7), (8), and 
n n 

(9), n having the meaning given in the Note in § 379 a, 

which see ; and proceed accordingly]. 



381a. Exaggeration of Vertical Dimensions of Both Space and 
Force Diagrams. — In case, as often happens, the axis of the 
given rib is quite a flat curve, it is more accurate (for find- 
ing M) to proceed as follows : 

After drawing the curve in its true proportions and pass- 
ng a vertical through the middle of each of the equal 
z/s's, compute the ordinate (y) of each of these middle points 
from the equation of the curve, and multiply each y by 
four (say). These quadruple ordinates are then laid off 
from the span upward, each in its proper vertical. Also 
multiply each load, of the given loading, by four, and then 
with these quadruple loads and quadruple ordinates, and 
the upper extremities of the latter as points in an exagge- 
rated rib-curve, proceed to construct a special equilibrium 
polygon, and the corresponding force diagram by the 
proper method ( for Class B, C, or D, as the case may be) 
for this exaggerated rib-curve. 

The moment, Hz, thus found for any section of the ex- 



ARCH-RIBS. 471 

aggerated rib-curve, is to be divided by four to obtain the 
moment in the real rib, in the same vertical line. To find 
the thrust and shear, however, for sections of the real rib, 
besides employing tangents and normals of the real rib W9 
must draw, and use, another force diagram, obtained from 
the one already drawn (for the exaggerated rib) by re- 
ducing its vertical dimensions (only), in the ratio of four 
to one. [Of course, any other convenient number besides 
four, may be adopted throughout.] 

382. Stress Diagrams. — Take an arch -rib of Class D t § 378, 
i.e., of fixed ends, and suppose that for a given loading (in- 
cluding its own weight) the special &%%%% .thrust 
equil. polygon and its force diagram 

have been drawn [§ 381]. It is re- *lfffff|f| * — -~ c oup»-e 
quired to indicate graphically the 
variation of the three stress-elements 
for any section of the rib, viz., the 
thrust, shear, and mom. of stress- 
couple, i" is constant. If at any 
point m of the rib a section is made, then the stresses in 
that section are classified into three sets (Fig. 432). (See 
§§ 295 and 367) and from § 367 eq. (3) we see that the ver- 
tical intercepts between the rib and the special equil. 
polygon being proportional to the products Hz or 
moments of the stress-couples in the corresponding sec- 
tions form a moment diagram, on inspection of which we 

can trace the change in this moment, Hz = ^~ , and 

e 

hence the variation of the stress per square inch, p 2 , (as 
due to stress couple alone) in the outermost fibre of any 
section (tension or compression) at distance e from the 
gravity axis of the section), from section to section along 
the rib. 

By drawing through lines On' and Of parallel re- 
spectively to the tangent and normal at any point m of the 
rib axis [see Fig. 433] and projecting upon them, in turn, 
the proper ray (B z in Fig. 433) (see eqs. 1 and 2 of § 367) 




Fig. 432. 



472 



MECHANICS OF ENGINEERING. 



we obtain the values of the thrust and shear for the sec- 
tion at m. When found in this way for a number of points 
along the rib their values may be laid off as vertical lines 
from a horizontal axis, in the verticals containing the re- 
spective points, and thus a thrust diagram and a shear dia- 
gram may be formed, as constructed in Fig. 433. Notice 
that where the moment is a maximum or minimum the 
shear changes sign (compare § 240), either gradually or 




Fig 



suddenly, according as the max. or min. occurs between 
two loads or in passing a load ; see m', e. g. 

Also it is evident, from the geometrical relations involv- 
ed, that at those points of the rib where the tangent-line 
is parallel to the " proper ray " of the force diagram, the 
thrust is a maximum (a local maximum) the moment (of 



AECH-EIBS. 473 

stress couple) is either a maximum or a minimum and the 
shear is zero. 

From the moment, Hz = &*,p 2 ^ ^ 

e 1 

tfiTll^t 

may be computed. From the thrust = Fp lt p x — — — — , (F 

F 

= area of cross-section) may be computed. Hence the 
greatest compression per sq. inch (pi-\-p 2 ) may be found in 
each section. A separate stress-diagram might be con- 
structed for this quantity (^1+^2). Its max. value (after 
adding the stress due to change of temperature, or to rib- 
shortening, for ribs of less than three hinges), wherever it 
occurs in the rib, must be made safe by proper designing 
of the rib. The maximum shear J m can be used as in §256 
to determine thickness of web, if the section is I-shaped, 
or box-shaped. See § 295. 



383. Temperature Stresses. — In an ordinary bridge truss 
and straight horizontal girders, free to expand or contract 
longitudinally, and in Classes A and B of § 378 of arch- 
ribs, there are no stresses induced by change of tempera- 
ture ; for the form of the beam or truss is under no 
constraint from the manner of support ; but with the arch- 
rib of two hinges (hinged ends, Class C) and of fixed ends 
(Class D) having immovable piers which constrain the dis- 
tance between the two ends to remain the same at all tem- 
peratures, stresses called " temperature stresses " are in- 
duced in the rib whenever the temperature, t, is not the 
same as that, t u , when the rib was put in place. These 
may be determined, as follows, as if they were the only 
ones, and then combined, algebraically, with those due to 
the loading. 

384. Temperature Stresses in the Arch-Rib of Hinged Ends, — 
(Class C, § 378.) Fig. 434. Let E and /be constant, with 



474 



MECHANICS OF ENGINEERING. 




Fig. 434. 



n , other postulates as in § 379. 
Let t — temperature of 
erection, and i = any other 
temperature ; also let I = 
length of span = OB (in- 
variable) and Tj= co-efficient 
of linear expansion of the 
material of the curved beam or rib (see § 199). At tempera- 
ture t there must be a horizontal reaction H at each hinge 
to prevent expansion into the form O'B (dotted curve), 
which is the form natural to the rib for temperature t and 
without constraint. We may .*. consider the actual form 
OB as having resulted from the unstrained form O'B by 
displacing 0' to 0, i.e., producing a horizontal displace- 
ment O'O =1 (t-t n )r r 

But O'O = Ax (see §§ 373 and 374) ; (N.B. B's tangent 
has moved, but this does not affect Ax, if the axis X is 
horizontal, as here, coinciding with the span ;) and the 
ordinate y of any point m of the rib is identical with its 
z or intercept between it and the spec, equil. polygon, 
which here consists of one segment only, viz. : OB, Its 
force diagram consists of a single ray X n' ; see Fig. 434. 
Now (§ 373) 



Ax = mf Mydi 



and 31= Hz = in this case, Hy 



l(t-t ) V : 



Flfo rdS; 



hence for graphics, and 
equal Js's, we have 



Ellit-Qr^H As 11 f . . . . (1) 

From eq. (1) we determine H t having divided the rib-curve 
into from twelve to twenty equal parts each called As . 

For instance, for wrought iron, t and t oi being expressed 
in Fahrenheit degrees, rj = 0.0000066. If E is expressed 
in lbs. per square inch, all linear quantities should be in 
inches and if will be obtained in pounds. 

I^y 2 may be obtained by § 375, or may be computed. R 
being known, we find the moment of stress-couple = Hy t 



ARCH-RIBS. 



475 



at any section, while the thrust and shear at that section 
are the projections of H, i.e., of x n' upon the tangent and 
normal. The stresses due to these may then be determined 
in any section, as already so frequently explained, and 
then combined with those due to loading. 

385. Temperature Stresses in the Arch-Ribs with Fixed Ends.— 
See Fig. 435. (Same postulates as to symmetry, E and 1 
constant, etc., as in § 380.) t and t have the same meaning 
as in § 384. 

Here, as before, we 
consider the rib to 
have reached its ac- 
tual form under tem- 
perature t by having 
had its span forcibly 
shortened from the 
length natural to 
temp, t, viz. : O'B', 
to the actual length OB, which the immovable piers compel 
it to assume. But here, since the tangents at and B are 
to be the same in direction under constraint as before, the two 
forces H, representing the action of the piers on the rib, 
must be considered as acting on imaginary rigid prolonga- 
tions at an unknown distance d above the span. To find 
H and d we need two equations. 

From § 373 we have, since M=Hz=H (y-d), 

TT B 

Ax,i.e.,Wd+BW,i.e.,l(t-Q v ,=^f o (y-d)yds . (2) 




Fig. 435. 



(3) 



or, graphically, with equal Js's 

EIl{t-t )r J =HAs\zZtf-dZ* y'\ . . 

Also, since there has been no change in the angle between 
end-tangents, we must have, from § 374, 

-—_rJMds=0 ; i.e., — jzds=0 ;i.e., f(y—d) ds=0 



476 MECHANICS OF ENGINEERING. 

or for graphics, with equal As\ I*y = nd . . . (4) 

in which n denotes the number of Js's. From (4) we 
determine d, and then from (3) can compute H. Drawing 
the horizontal F G, it is the special equilibrium polygon 
(of but one segment) and the moment of the stress-couple 
at any section = Hz, while the thrust and shear are the 
projections of E— x n' on the tangent and normal respect- 
ively of any point m of rib. 

For example, in one span, of 550 feet, of the St. Louis 
Bridge, having a rise of 55 feet and fixed at the ends, the 
force H of Fig. 435 is = 108 tons, when the temperature is 
80° Fahr. higher than the temp, of erection, and the en- 
forced span is 3*^ inches shorter than the span natural to 
that higher temperature. Evidently, :.f the actual temp- 
erature t is lower than that t () , of erection, 17 must act in a 
direction opposite to that of Figs. 435 and 434, and the 
"thrust " in any section will be negative, i.e., a pull. 

386. Stresses Due to Rib-Shortening — In § 369, Fig. 407, the 
shortening of the element AE to a length A'E, due to the 
uniformly distributed thrust, PiF } was neglected as pro- 
ducing indirectly a change of curvature and form in the 
rib axis ; but such will be the case if the rib has less than 
three hinges. This change in the length of the different 
portions of the rib curve, may be treated as if it were due 
to a change of temperature. For example, from § 199 we 
see that a thrust of 50 tons coming upon a sectional area 
of F = 10 sq. inches in an iron rib, whose material has a 
modulus of elasticity = E = 30,000,000 lbs. per sq. inch, 
and a coefficient of expansion rj = .0000066 per degree 
Fahrenheit, produces a shortening equal to that due to a 
fall of temperature (t —t) derived as follows : (See § 199) 
(units, inch and pound) 

(t -t)= P = 100 ' 000 -50° 

Vo ; FE-q 10 x 30,000,000 x.0000066~ 

Fahrenheit. 

Practically, then, since most metal arch bridges of 

classes C and D are rather flat in curvature, and the thrusts 



ABCH-EIBS. 477 

due to ordinary modes of loading do not vary more than 20 
or 30 per cent, from each other along the rib, an imagin- 
ary fall of temperature corresponding to an average thrust 
in any case of loading may be made the basis of a con- 
struction similar to that in § 384 or § 385 (according as the 
ends are hinged, ox fixed) from which new thrusts, shears, 
and stress-couple moments, may be derived to be combin- 
ed with those previously obtained for loading and for 
change of temperature. 

387. Resume — It is now seen how the stresses per square 
inch, both shearing and compression (or tension) may be 
obtained in all parts of any section of a solid arch-rib or 
curved beam of the kinds described, by combining the re- 
sults due to the three separate causes, viz.: the load, 
change of temperature, and rib-shortening caused by the 
thrusts due to the load (the latter agencies, however, com- 
ing into consideration only in classes G and D, see § 378). 
That is, in any cross-section, the stress in the outer fibre 
is, [letting T h ', T h ", T h '" } denote tlfe thrusts due to the 
three causes, respectively, above mentioned ; (Hz)', (Hz)", 
{Hz)"\ the moments] 

= T * ± T *+ T *"± e S(Hz)' ± {Hz)"± (Hzy'l . . . (1) 

i.e., lbs. per sq. inch compression (if those units are used). 
The double signs provide for the cases 
where the stresses in the outer fibre, due 
to a single agency, may be tensile. Fig. 
436 shows the meaning of e (the same 
used heretofore) /is the moment of in- 
ertia of the section about the gravity 
axis (horizontal) G. F = area of cross- 
section. \e x — e\ cross section symmet- 
rical about C], For a given loading we 
may find the maximum stress in a given rib, or design the 
rib so that this maximum stress shall be safe for the ma- 
terial employed. Similarly, the resultant shear (total, not 




478 



MECHANICS OF ENGINEERING. 



per sq. inch) = J' ± J" ± J'" is obtained for any section 
to compute a proper thickness of web, spacing of rivets, 
etc. 

388 The Arch-Truss, or braced arch. An open-work 
truss, if of homogeneous design from end to end, may be 
treated as a beam of constant section and constant moment 
of inertia, and if curved, like the St. Louis Bridge and the 
Coblenz Bridge (see § 378, Class D), may be treated as an 
arch-rib.* The moment of inertia may be taken as 



/-■«(*)• 



where F i is the sectional area of one of the pieces ] to the 
curved axis midway between them, Fig. 437, and h = dis- 
tance between them. 



Jjv^ 




Fig. 438. 



Fig. 437. 



Treating this curved axis as an arch-rib, in the usual 
way (see preceding articles), we obtain the spec. equiL pol. 
and its force diagram for given loading. Any plane "1 to 
the rib-axis, where it crosses the middle m of a " web- 
member," cuts three pieces, A, B and C, the total Com- 



pile St Louis Bridge is not strictly of constant moment of inertia, being somewhat 
strengthened near each pier 



AECH-E1BS. 



479 



pressions (or tensions) in which are thus found : For the 
point m, of rib-axis, there is a certain moment = Hz, a 
thrust = T h , and a shear = J, obtained as previously ex- 
plained. We may then write P sin /? = J (1) 

and thus determine whether P is a tension or compres- 
sion ; then putting P'+P" ± P cos /3 = T h 2 

(in which P is taken with a plus sign if a compression, and 
minus if tension), and 



(P>-P")" = Hz 



(3) 



we compute P* and P", which are assumed to be both com- 
pressions here. /? is the angle between the web member 
and the tangent to rib-axis at m, the middle of the piece. 
See Fig. 406, as an explanation of the method just 
adopted. 



HORIZONTAL, STRAIGHT GIRDERS. 

389. Ends Free to Turn. — This corresponds to an arch- 
rib with hinged ends, but it must be understood that there 
is no hindrance to horizontal motion. (Fig. 439.) In 




Fig 439. 



treating a straight beam, slightly bent under vertical forces 
only (as in this case with no horizontal constraint), as a 



480 MECHANICS OF ENGINEERING. 

particular case of an arch-rib, it is evident that since the 
pole distance must be zero, the special equil. polygon will 
have all its segments vertical, and the corresponding force 
diagram reduces to a single vertical line (the load line). 
The first and last segments must pass through A and B 
(points of no moment) respectively, but being vertical will 
not intersect P i and P 2 ; i.e., the remainder of the special 
equilibrium polygon lies at an infinite distance above the 
span AB. Hence the actual spec, equil. pol. is useless. 

However> knowing that the shear, J, and the moment 
M (of stress couple) are the only quantities pertaining to 
any section m (Fig. 439) which we wish to determine (since 
there is no thrust along the beam), and knowing that an 
imaginary force H' , applied horizontally at each end of the 
beam, would have no influence in determining the shear 
and moment at m as due to the new system of forces, we 
may therefore obtain the shears and moments graphically 
from this new system (viz.: the loads P l} etc., the vertical 
reactions Fand V n , and the two equal and opposite i?"'s). 
[Evidently, since H 1 ' has no moment about the neutral 
axis (or gravity axis here), of m, the moment at m will be 
unaffected by it ; and since H" has no component ~| to the 
beam at m, the shear at m is the same in the new system 
of forces, as in the old, before the introduction of the 
ZT's.] 

Hence, lay off the load-line 1 . . 2. . 3, Fig. 439, and con- 
struct an equil. polyg. which shall pass through A and B 
and have any convenient arbitrary H" (force) as a pole 
distance. This is done by first determining n on the load- 
line, using the auxiliary polygon Aa'F, to a pole 0' (arbi- 
trary), and drawing O'n' || to A'B'. Taking 1 ' on a hori- 
zontal through n\ making 0'n'=H", we complete the 
force diagram, and equil. pol. AaB. Then, z being the ver- 
tical intercept between m and the equil. polygon, we have: 
Moment at m~M=H"z (or=iZV also), and shear at m, or 
J, =2 . . n' y i.e., = projection of the proper ray i? 2 > or 
0" . . 2, upon the vertical through m. Similarly we ob- 
tain M and J at any other section for the given load. (See 



ARCH-RIBS ; SPECIAL CASE ; STRAIGHT. 



481 



§§ 329, 337 and 367). The moment of inertia need not be 
constant in this case. 

390, Straight Horizontal Prismatic Girder of Fixed Ends at Same 
Level. — No horizontal constraint, hence no thrust. / con- 
stant. Ends, at same level, with end-tangents horizontal. 
We may consider the whole beam free (cutting close to the 
walls) putting in the unknown upward shears J and J u3 




and the two stress couples of unknown moments M and 
M n at these end sections. Also, as in § 388, an arbitrary 
H" horizontal and in line of beam at each extremity. Now 
(See Fig. 33) the couple at and the force H" are equiv- 
alent to a single horizontal H" at an unknown vertical dis- 
tance c below ; similarly at the right hand end. The 
special polygon FG is to be determined for this new sys- 
tem, since the moment and shear will be the same at any 
section under this new system as under the real system. 
The conditions for determining it are as follows : Since 
the end-tangents are fixed, IMAs =0 .*. 2*zAs=0 and since 



482 MECHANICS OF ENGINEERING. 

O's displacement relatively to B's tangent is zero we Lave 
IMxAs =0 .\ 2H"sxA8=0 •'. SxzAa =0. See § 374. Hence 
for Equal ^tfs's, ±'(z)=0 and I{xz)—0. Now for any pole 0'" 
draw an equil. pol. F"'G'" and in it (by § 377; see Note) 
locate v'"m'" so as to make J{s"')=0 and 2 , (cc«' // )=0. 
Draw verticals through the intersections E'" and L'", to 
determine E and Z on the beam, these are the points of 
inflection (i.e., of zero moment), and are points in the re- 
quired special polygon EG. 

Draw 0"V || to v'"m'" to fix n". Take a pole 0" on 
the horizontal through n", making G*'yi"=H" (arbitrary), 
draw the force diagram 0" 1234 and a corresponding 
equilibrium polygon beginning at E. It should cut X, 
and will fulfil the two requirments 2*(z)=0 and 2%(xz)=0, 
with reference to the axis of the beam O'B'. The moment of 
the stress-couple at any section m will be M—R n z, and the 
shear J = the projection of the " proper ray " of the force 
diagram 0" . . 1, 2, etc., upon the vertical (not in the trial 
diagram 0'". . 1, 2, etc.). As far as the moment is concern- 
ed the trial polygon F"' G"' will serve as well as the special 
polygon EG ; i.e., M=W"z"' as well as H"z, H" f being the 
pole-distance of 0"' ; but for the shear we must use the 
rays of the final and not the trial diagram. 

The peculiarity of this treatment of straight beams, 
considered as a particular case of curved beams, consists 
in the substitution of an imaginary system of forces in- 
volving the two equal and opposite, and arbitrary H's, for 
the real system in which there is no horizontal force and 
consequently no " special equilibrium polygon," and thus 
determining all that is desired, i.e., the moment and shear 
at any section. 

In the polygon EG the student will recognize the " mo- 
ment-diagram " of the problems in Chaps. Ill and IV. 

He will also see why the shear is proportional to the 

slope — — of the moment curve in those chapters. For 
ax 

example, the " slope " of the second segment of the poly- 
gon EG, that segment being || to 0" 2, is 



ABCH-BIBS; SPECIAL CASE; STEAIGHT. 483 



tang, of angle 20'W'=2w"-*-0"w"=shear ^ H" 

and similarly for any other segment ; i.e., the tangent of 
the inclination of the " moment curve," or line, is propor- 
tional to the shear. 

It is also interesting to notice with the present problem 
of a straight beam, that in the conditions 

I(zJs)=0 and 2(zJs)x=0, 

for locating the polygon FG, each ds is T to its 2, and 
that consequently each zAs is the area of a small vertical 
strip of area between the beam and the polygon, and 
[zAs)x is the " moment" of this strip of area, about 0' the 
origin of x. Hence these conditions imply ; first, that the 
area EWL between the polygon and the axis of the beam 
on one side is equal to that (0'FE-\-LB' G) on the other 
&ide, and, secondly, that the centre of gravity of EWL lies 
in the same vertical as that of O'FE and LB 1 G combined. 
Another way of stating the same thing is that, if we join 
FG, the area of the trapezoid FO'B' G is equal to that of the 
figure FEWLG, and their centres of gravity lie in the same 
vertical. A corresponding statement may be made (if we join 
F'"G'") for the trapezoid F"'v'"m'"G"> and figure 
F>»E'"W"'L'"G"\ 



484 



MECHANICS OF EXGLNTEEBTXG. 



CHAPTEK XII. 



GRAPHICS OF CONTINUOUS GIRDERS. 



[Mainly due to Prof. Mohr : of Aix-la-Chapelle ] 



;« &%— 4 



391. The Elastic Curve of a Horizontal Loaded Beam, Homoge* 
neous and Originally Straight and Prismatic, is an Equilibrium 
Polygon, whose "load-line" is vertical \ 
and consists of the successive products \ 
Mdx [treated as if they were loads A 
each applied through the middle of 
its proper dx\ and whose "pole dis- 
tance" is EI. Fig 441 (exaggerated). 

Let AO and 00 be any two con- 
secutive equal elements of a very 
flat elastic curve (as above described). 
Prolong AO to cut NO. Then from 
§ 231, eq. (7), we have 




d?y_{Mdx) 



dx 



EI 



(1) 



where d 2 y=DO;&nd > hence, if a triangle (Fig. 442) O'D'C, 
be formed with 0'B\ II to OD, O'C II to 00, and D'C ver- 
tical, while its (horizontal) altitude O'n is made equal, by 
scale, to EI of the beam, then from the similarity of the 
triangle 0D0 and O'D'C and the proportion in eq. (1) we 
see that D'O' must represent the product Mdx on the same 
scale by which O'n represents EI. M is the moment of 



CONTINUOUS GIRDER BY GRAPHICS. 485 

the stress-couple at the section whose neutral axis is pro- 
jected in 0. 

Similarly, if 31' is the moment of the stress -couple at 
(7, and we draw O'F', \\ to CF, C'F' must represent M'dx 
(on same scale). It is therefore apparent that the line 
AOCF bears to the figure O'D'C'F' the same relation 
which an equilibrium polygon (for vertical forces) does to 
its force diagram, the " loads " of the force-diagram being 
the successive values of 3Idx laid off to scale, while its 
" pole-distance " is EI laid off on the same scale. [As if 
Mdx and M'dx were loads suspended at and G respec- 
tively.] 

Practically, since any actual elastic curve is very flat, 
and since a change of pole-distance will change all verti- 
cal dimensions of the equilibrium polygon in an inverse 
equal ratio, we may exaggerate the vertical dimensions of 
the elastic curve by choosing a pole distance smaller than 
EI in any convenient ratio, n. Any deflection in the elas- 
tic curve thus obtained will be greater than its true value 
in the same ratio n. 

Graphically, in order to draw exaggerated elastic curves 
according to this principle, we obtain approximate results 
by dividing the length of the beam into a number of equal 
Ax's, draw verticals through the middles of the Ax's, as 
" force-verticals,"and lay off as a " load-line " to any con- 
venient scale the corresponding values of 31 Ax in their 
proper order. 

The quality of the product 31 Ax is evidently (length) 2 x 
force, and with the foot and pound as units such a product 
may be called so many (sq. ft.) (lbs.). It will be noticed 
that these products (31Ax) are proportional to, and maybe 
represented by, the areas of the corresponding vertical 
strips of the " moment-diagram " proper to the case in 
hand, These strips together make up the " moment-area" 
as it may be called, lying between the moment curve and 
its horizontal axis (which is the axis of the beam itself, 
according to §§ 389 and 390). 



48G MECHANICS OF ENGINEERING. 

392, Mohr's Theorem. — The principle of the previous 
paragraph may therefore be enunciated as follows : That 
just as the moment curve (of a straight prismatic horizontal 
beam) between tioo consecutive supports is an equilibrium poly- 
gon for the loading between those supports, so also is the elastic 
curve itself an equilibrium polygon for the " moment-area " con- 
sidered as a loading. 

In dealing with the moment-curve of a single span the 
pole distance is arbitrary (§§ 389 and 390), but the position 
of the pole relatively to the load line in other respects, and 
the location of the moment-curve (equil.-pol.) relatively to 
the beam (considered to be still straight for this purpose), 
depend on whether the beam simply rests on the two sup- 
ports, without projecting beyond : or is built in, and at 
what angles ; or as with a continuous girder y on the inclina- 
tion of the tangent-lines at the supports, as influenced by 
the presence of loads on all the spans, and on whether all 
supports are on the same level or not. 

For example, in § 389, for a single span, the ends of 
beam being simply supported without overhanging, the 
pole 0" must be on a horizontal through n', and the mo- 
ment curve must pass through the extremities A and B of 
the beam, thus giving a " moment-area " lying entirely on 
one side of the beam (or axis from which the moment or- 
dinates, z, are to be measured) ; whereas, in § 390, also a 
single span, where the ends of the beam are built in hori- 
zontally and at the same level, the pole must be taken on 
the horizontal through w", and the moment-curve FEWLG 
must intersect the beam in the points E and L (E, L, and 
n fP being found as prescribed in that problem), and thus 
lies partly above and partly below the beam. It will be 
necessary, later, to distinguish the upper and lower parts 
of the moment-area as positive and negative. 

In drawing the equilibrium polygon which constitutes 
the actual elastic curve, however, and hence making use of 
the successive small vertical strips of the moment-area, 
(when found) as if they were loads, to form a load-line ac- 
cording to a convenient scale, the pole distance is not ar- 



CONTINUOUS GIRDER BY GRAPHICS. 



487 



bitrary but must be = EI on the same scale. Still, since 
for convenience we must always greatly exaggerate the 
vertical scale of the elastic curve, we may make the pole 
distance = EI-^n and thus obtain an elastic curve whose 
vertical dimensions are n times as large as those of the real 
curve ; while the position of the pole will depend in the 
direction of the tangent lines at the extremities of the span. 
An example will now be given. 

393 Example of an Elastic Curve (Beam Prismatic) Drawn as 
an Equilibrium Polygon Supporting the Moment-Area as Loading. 
— Let the beam be simply supported at its extremities (at 
the same level), and bear a single eccentric load P, Fig. 
443, its own weight being neglected. The moment-area 




Fig. 443. 



consists of a triangle ACB [see first part of § 260, or use 

the graphic method of § 389, thus utilizing a force diagram 

012.], its altitude being the moment represented by the 

PI I 
ordinate CD and having a value = — J-?. Hence the total 

c 



moment- area = y 2 base AB x mom. CD 

i.e., = ^2 X y 2 i=y 2 PUz 



488 MECHANICS OF ENGINEERING. 

Divide AB into (say) eight equal Jx's (eight are rather 
few in practice ; sixteen or twenty would be better) and 
draw a vertical through the middle of each. Note the 
portion of each of these vertical intercepts between the 
axis of the beam and the moment-curve ACB. The pro- 
ducts M Ax for the different subdivisions are proportional 
to these intercepts, since all the Ax's are equal, and are the 
respective moment-areas of the Ax's. 

Treating these products as if they were loads, we lay off 
the corresponding intercepts (or their halves, or quarters, 
or other convenient fractional part or multiple), from E 
downwards to form a vertical " load-line," beginning with 
the left-hand intercept and continuing in proper order. 

As to what scale this implies, we determine by dividing 
the total moment-area thus laid off, viz. : j4 Pl\k, say in 
(sq. in.) (lbs.), by the length of EF in inches, thus obtain- 
ing the number of (sq. in.) (lbs.) which each linear inch of 
paper represents. 

On this scale the number of inches of paper required to 
represent the EI of the beam is so enormous, that in its 
stead we use the nth portion, n being an arbitrary abstract 
number of such magnitude as to make EI -f- n a con- 
venient pole-distance, TS. 

The proper position of the pole 0' on the vertical TW, 
is fixed by the fact that the elastic curve, beginning at A, 
must terminate in B, at the same level as A. Hence, 
assuming any trial pole as 0", and drawing rays in the 
usual manner (except that, as henceforth, the pole is taken 
on the right of the " load-line," instead of on the left, so 
as to make the resulting equilibrium polygon correspond 
in direction of curvature to the actual elastic curve), we 
draw the corresponding equilibrium polygon A"B". De- 
termining n' by drawing through 0" a line || to the right 
line A"0", we draw a horizontal through n' to intersect 
TW in 0', the required pole. 

With 0' as pole a new equilibrium polygon begun at A f 
will terminate in B f and its vertical ordinates will be n 
times as great as those of corresponding points on the 



CONTINUOUS GIRDER BY GRAPHICS. 489 

actual elastic curve AB. The same relation holds between 
the tangents of the angle of inclination to the horizontal 
at corresponding points (i.e., those in same vertical)of the 
two curves. 

394. Numerical Case of Foregoing Example. — With the inch 

and pound as units, let P = 120 lbs., l L = 40 in., l 2 = 80 in. 
while the prismatic beam is of timber having a modulus 
of elasticity E = 2,000,000 lbs. per sq. inch, and is rectan- 
gular in section, being 2 in. wide and 4 in. high, so that 
(its width being placed horizontally) the moment of iner- 
tia of the section is I = 1 /* e bh s = y i2 x2x64 = 10^ bi- 
quadratic inches (§ 90.) Kequired the maximum deflection. 
Adopting 1:20 as the scale for distances (i.e., one linear 
inch of paper to twenty inches of actual distance) we make 
the horizontal AB 6 in. long, Fig. 443, and AD 2 in., tak- 
ing a point C at convenience in the vertical through D, 
and joining AC and CB, thus determining the moment- 
diagram for this case. [As to what pole distance, H, is 
implied in this selection of (7, is immaterial in this simple 
case of a single load ; hence we do not draw the corre- 
sponding force-diagram at all.] We divide AB into eight 
equal parts and draw a vertical through the middle of 
each. The intercepts, in these verticals, between AB and 
the broken line ACB we lay off from E toward F as pre- 
scribed in § 393. (By taking DC small enough the line EF 
will not be inconveniently long.) 

Suppose this length EF measures 6.4 inches on the 
paper (as in the actual draft by the writer). Since it rep- 
resents a moment-area of 

^py 2 =^ x 120x40x80=192,000 (sq. in.) (lbs.), the scale 
of our "moment-area-diagram," as we may call it, must be 
192,000-^6.4=30,000 (sq. in.) (lbs.) per linear inch of paper. 

Now ^2=21,333,333 (sq. in.) (lbs.), which on the above 
scale would be represented by 711 linear inches of paper. 
With ?z=100, however, Ave lay off ST=EI+n = 7. 11 inches 
of paper as a pole distance, and with a trial pole 0" in 



490 MECHANICS OF ENGINEERING. 

the vertical TW draw the trial equilibrium polygon or 
elastic curve A"B", and with it determine n', then the final 
polygon A'B' as already prescribed. In A'B' we find the 
greatest ordinate, NK, to measure 0.88 inches of paper, 
which represents an actual distance of 0.88 x 20 = 17.6 
inches But the vertical dimensions of the exaggerated 
elastic curve A'B' are w=100 times as great as those of the 
actual, hence the actual max. deflection is d=17.6-^-n=0.176 
in. [This maximum deflection could also be obtained from 
the oblique polygon A"B" whose vertical dimensions are 
equal to those of A'B', By the formula of § 235 we ob- 
tain d=0.174: inches.] 

395, Direction of End-Tangents of Elastic Curve in the Foregoing 
Problem. — As an illustration bearing on subsequent work 
let us suppose that the only result required in § 394 is 
tan a , i.e., the tangent of the angle B'A'T', which the 
tangent-line B'T' to the elastic curve at the extremity A', 
Fig. 343, makes with the horizontal line A'B' , (tan. a is 
called the "slope" at A.) ~LetB'S' be the tangent-line at 
B'. These two " end-tangents " are parallel respectively to 
EO' and FO', and intersect at some point B. Now since 
A' KB' is an equilibrium polygon sustaining an imaginary 
set of loads represented by the successive vertical strips of 
the moment-area ACB, the intersection R must lie in the 
vertical containing the centre of gravity, U, of that mo- 
ment-area [§ 336 1 . 

Hence, if the vertical containing 27 is known in advance^ 
or, as in the present case, is easily constructed without 
making the strip-subdivision of § 394, we may determine 
the end-tangents very briefly by considering the whole 
moment-area, M.A., (considered as a load) applied in the 
vertical through U, as follows : 

Since ACB is a triangle, we find U by bisecting AB in 
X, joining GX f and making XJJ = % XC, and then draw a 
vertical through U. Laying off EF=6A inches [so as to 
represent a moment-area of 192,000 (sq. in.) (lbs.) on a 
scale of 30,000 (sq. in.) (lbs.) per linear inch of paper] > 



CONTINUOUS GIRDER BY GRAPHICS. 491 

and making ST=7.11 inches as before, we assume a trial 
pole 0" on TW y draw the two rays 0"E and 0"F, construct 
the corresponding trial polygon of two segments A"R"B"> 
for the purpose of finding ri '. With a pole 0' on TJFand 
on a level with n' we draw the two rays O'E and O'E, and 
the corresponding segments A'R S and RB'. LB' should be 
on a level with A', as a check.) These two segments are 
the end-tangents required. 
We have, therefore, 



tan a = B'T' •*- A'B' = B" T" 4- A'B' 

In the present numerical problem we find B'T' to mea- 
sure 3 in. of paper, i.e., 60 in. of actual distance for the 
exagg. elastic curve, and therefore 0.60 in. in the real elas- 
tic curve (with n = 100) 

, 0.60 in. A AAcr 

.-. tan. a = — — — = 0.005 
120 in. 

It is now evident that the position and direction of the 
end-tangents of the elastic curve lying between any two sup- 
ports are independent of the mode of distribution of the 
moment-area so long as the amount of that moment-area and 
the position of its centre of gravity remain unchanged. This 
relation is to be of great service. 

396. Re-Arrangement of the Moment- Area. — As another illus- 
tration conducing to clearness in later constructions, let 
us determine by still another method the end-tangents of 
the beam of §§ 394 and 395. See Fig. AAA. As already 
seen, their location is independent of the arrangement of 
the moment-area between. Let us re-arrange this moment- 
area, viz., the triangle AGB, in the following manner : 

By drawing AX parallel to BC, and prolonging BC to V 
in the vertical through A, we may consider the original 
moment-area AGB to be compounded of the positive mom.- 
area VBXA, a parallelogram, with its gravity-vertical 
passing through D, the middle of the span ; of the negative 
mom. -area VCA, sl triangle whose gravity-vertical passes 



492 



MECHANICS OF ENGINEERING. 



through D x making AD L =% AD; and of another negative 
mom. -area, the triangle ABX, whose gravity-vertical passes 
through D 3 at one-third the span from B. That is, the 
(ideal) positive load ACB is the resultant of the positive 
load (M.A.) 2 and the two negative loads (or upward pulls) 
■(M.A.\ and (M.A.\ and may therefore be replaced by them 
without affecting the location of the end-tangents, at A 



, k-40-in 




and B y of the elastic curve AB. These three moment- 
areas are represented by arrows, properly directed, in the 
figure, but must not be confused with the actual loads on 
the beam (of which, here, there is but one, viz., P). 

From the given shapes and dimensions, since ACB = 
192,000 (sq. in.) (lbs.), we easily derive by geometrical 
principles : 

•(3LA.) 2 = +576,000 (sq. in.) (lbs.) 
\M.A.\= -'96,000 

(M.A.) S = -288,000 

Hence, with a pole distance EI -*• n = 7.11 in. as before, 
and a " moment-load-line " formed of 1'2' = (M.A.) ly (on 
scale of 30,000 (sq. in. lbs.) to one inch) 2'3' = (M.A.%, and 
3'4' — (M.A.\, first with a trial pole 0", construct the tria^ 



CONTINUOUS GIRDER BY GRAPHICS. 493 

polygon A"B'\ and find n' in usual way (§ 337) ; then take 
a pole 0' on the horizontal through n' and the vertical 
TW 3 and draw the new polygon A' 123 B'. It should 
pass through B' on a level with A\ and A'\ and B'3 are the 
required end-tangents (of the exagg. elastic curve). 

[Note — If B' were not at the same level as A', but 
(say) 0.40 in. below it, B f should be placed at m, a distance 

• x = 2 inches (on the paper) below its present posi- 

tion, (since the distance scale is 1:20 and n = 100, in this 
case) and the " abutment -line " of final polygon would be 
A'm\ 

Of course, this special re-arrangement of the moment- 
area is quite superfluous in the present problem of a dis- 
continuous girder (not built in), but considerations of this 
kind will be found indispensable with the successive spans 
of a continuous girder. 

387. Positive and Negative Moment- Areas in Each Span of a 
Continuous Girder (Prismatic). — In the foregoing problem of a 
discontinuous girder (covering one span only) not built in at 
the ends (otherwise it would be classed among continuous 
girders), the moment-curve, or equilibrium polygon of 
arbitrary H 9 is easily found by § 389 without the aid of the 
elastic curve, and the end moments are both zero ; (i.e., the 
moment-curve meets the beam in the end- verticals) but in 
each span of a continuous girder the end-moments are not 
zero (necessarily), and the points in the end-verticals where 
the moment-curve must terminate (for an assumed H) can 
not be found without the use of the elastic curve (or of 
some of its tangents) of the whole beam, dependent, as it is, 
upon the loading on ail the spans, and the heights of the 
supports. 

Let Fig. 445 show, in general, any one span of a pris- 
matic continuous girder {prismatic; hence 7 is constant), 
between two consecutive supports A and B . P u P 2 , etc., 
are the loads on the span. 

[If the displacement of A n relatively to the end-tangent 
at B,,, and the angle between the end-tangents (of elastic 



494 



MECHANICS OF ENGINEERING. 



curve) were known, the. moment-curve or equilibrium 
polygon FWG, (AB being the axis of beam) might be 
found by a process similar to that in § 390, but the elastic 
curves in successive spans are so inter-dependent that the 
above elements can not be found directly.] 



^rr^ 




Fio. 445. 

We now suppose, for the sake of discussion, that the 
whole girder has been investigated (by a process to be 
presented) for the given loads, spans, positions of supports, 
etc., and then the moment-curve FEWLG found, with the 
corresponding force-diagram, for the span in the figure 
and some arbitrary H. The horizontal line AB represents 
the axis of the beam (for this purpose considered straight 
and horizontal) as an axis from which to measure the 
moment ordinates. 

Thus, the moment (of the stress-couple) at A is = H X 
AF; at B, Hx BG ;at E andZ, zero (points of inflection). 

Now, according to the usual conceptions of analytical 
geometry, we may consider the portion EWL, of the mo- 
ment-area, above AB as positive, and those below, AEF 
and LB G, as negative ; but since not one of these three 



CONTIGUOUS GIRDER BY GRAPHICS. 495 

areas, nor the position of its gravity -vertical, is known in 
advance, since they are not independent of the other 
spans, a more advantageous re-arrangement of the moment- 
area may be made thus : 

Join FG and FB, and we may consider the original 
moment-area replaced by the following three component 
areas: the positive moment-area FEWLGF (shaded by ver- 
tical lines) ; the negative triangular moment-area AFB ; 
and the negative triangular moment-area BFG (the nega- 
tive moment-areas being shaded by horizontal lines). (In 
subsequent paragraphs, by positive and negative moment- 
areas will be implied those just mentioned.) 

These three moment-areas, treated as loads, each applied 
in its own gravity -vertical, and considered in any order, 
may be used instead of the real distributed moment-area, 
as far as determining, or dealing with, the end-tangents of the 
elastic-curve at A and B is concerned (§ 396), and the fol- 
lowing advantages will have been gained : 

(1.) The amount of the positive moment-area, (M.A.\ in 
Fig. 445, (depending on the area lying between the polygon 
FEW G and the abutment-line FG of the latter), and the 
position of its gravity -vertical, are independent of other spans, 
and can be easily found in advance, since this moment-area 
and gravity -vertical are the same as if the part of the beam 
covering this span were discontinuous and simply rested on the 
supports A and B lt , as in § 389. 

(2.) The gravity-vertical of the left-hand negative mo- 
ment-area, (M.A.\, is always one-third the span from the 
left end-vertical, A A'\ that of the other, (M.A.) 3i an equal 
distance from the right end-vertical, B B'. 

(3.) The two (right and left) negative moment-areas 
are triangular, each having the whole span I for its alti- 
tude, and for its base the intercept AF {ox BG) on which 
the end-moment depends. Hence, if the amounts of these 
negative moment-areas have been found in any span, we 
may compute the values which AF and BG must have for 
a given H, and thus determine the terminal points F and 
G of the moment-curve of that span (for that value of H), 



496 MECHANICS OF ENGINEERING. 

For example, if (M.A.) L has been found (by a process not 
yet given) to be 160,000 (sq. in.) (lbs.) while AB = I = 160 
in., then the moment 3I A which Misrepresents, is computed 
from the relation 

(M.A.\ = y 2 ~AB x M A 

or, if A = 2xi y =2000. in. lbs. 
160 

If ZThas been chosen = 100 lbs. we put HxAF = 2000 
and obtain Mi* 7 = 20 inches of actual distance, so that with 
a scale of 1:20 for distances AF would be one linear inch 
of paper. (Of course, in computing BG the same value of 
H must be used.) With H= 100 lbs., then, and F and G 
as known points of the equilibrium polygon FEW G, it 
is easily drawn by the principles of § 341. 

We thus notice that the amounts of the two negative 
moment-areas are the only elements affected by the con- 
tinuity of the girder, in this re-arrangement of the actual 
moment-areas. 

In the lower part of Fig. 445 A' and B f represent the 
extremities of the (exagg.) elastic curve, the vertical dis- 
tance B'B'", of B' from the horizontal through A' (in case 
the two supports A and B Q are not at same level, as we 
here suppose for illustration) being laid off in accordance 
with the principles of the note in § 396. 

Note. — It is now evident that if the "false polygon" (as 
it will be called) A'123B' has been obtained (and means for 
doing this will be given later) in which the first and last 
segments are the end tangents of the (exagg.) elastic curve, 
and which bears the same relation to the three moment 
areas just mentioned, as that illustrated in Fig. 444, we 
may proceed further to determine the amounts of {M.A.\ 
and (M.A.\ as follows, by completing the moment-area dia- 
gram : 

Having laid off the known (M.A.) 2 (or positive moment- 
area) = 2'3', and 8 T—EI^n, a line parallel to 12 drawn 
through 2', determines the pole 0', through which paral-= 



CONTINUOUS GJRDEE BY GRAPHICS. 497 

leis to A'l and B'S will fix 1' and 4' on the vertical SV 
and thus determine 1'2'=(M.A.\ and 3'±'=(M.A.) 3 . Their 
numerical values are then computed in accordance with 
the scale of the moment-area diagram. 

The polygon ^'1235' will be called the "false polygon " 
of the span m question, its end-segments being the end- 
tangents of the elastic curve. 



Values of the Positive Moment-Area in Special Cases - 

*or several special cases these are easily computed, and 
as an illustration, Fig. 446 shows a continuous girder AF 



uimu 




Fig. 446. 



of five spans, all six supports on a level, and the weight of 
the beam neglected. At the extremities A and F, as at 
the other supports, the beam is not built in, but simplv 
touches each support in one point; hence the moments at 
A and ^are zero, i.e., the moment curve must pass through 
A and F, so that in the first span the left negative mo- 
ment-area, and in the fifth span the right negative mo- 
ment-area, are zero. The positive moment-areas are 
shaded. 

On the first span is placed a uniformly distributed load 
W over the whole span V. ... the positive moment-area 

mf^f 8 Same aSinthe case of Fi e- 235 [see 
WMJ7] and being represented by a parabolic segment 
whose area is two-thirds that of the circumscribing rec- 
tangle, its value is R 



498 MECHANICS OF ENGINEERING. 

(M.A.y 2 =2/ 3 .y 8 wi'xi'= l / l2 wp . . (i) 

[see eq. (2) § 242], while its gravity- vertical bisects the 
span. 

The only load on the second span is a concentrated one, 
P' , at distances £/' and l 2 from the extremities of the span; 
hence the positive moment-area is triangular and has a 
value 

(M.A.) 2 "=y 2 P"l l "l 2 » . . (2) 

as in § 393. Its gravity vertical may easily be constructed 
as in Fig. 443 [see (1) § 397]. 

The third span carries no load; hence its positive mo- 
ment-area is zero, and the actual moment-area is composed 
solely of the two triangular negative moment-areas CDH 
and DHI, the moment-curve consisting of the single 
straight line HI. 

The fourth span carries a uniform load W IY =w IY l 1Y , and 
.*. has a positive moment-area 

(M.A.)?='/ l2 wi\ry . . . (3) 

as in eq. (1), acting through the middle of the span (grav- 
ity-vertical). 

Since the fifth and last span carries no load, its positive 
moment-area is zero, the moment curve being the straight 
line JF, so that the actual moment-area is composed of 
the left-hand negative moment-area. 

At F it is noticeable that the reaction or pressure of the 
support must be from above downward to prevent the 
beam from leaving the point F; i.e., the beam must be 
" latched down," and the reaction is negative. 

If the beam were built in at A (or F) the moment at 
that section would not be zero, hence the left (or right) 
neg. moment-area would not be zero in that span, as in 
our present figure. But in such a case the tangent of the 
elastic curve would have a 'known direction at A (or F) and 
the problem would still be determinate as will be seem 






CONTINUOUS GIRDER BY GRAPHICS. 499 

A' . ♦ . F' gives an approximate idea (exaggerated) of 
the form of the elastic curve of the entire girder. A change 
in the loading on any span would affect the form of this 
curve throughout its whole length as well as of all the 
moment curves. 

Note. — It is important to remark that any two of the 
triangular negative moment-areas which have a common 
base (hence lying in adjacent spans) are proportional to 
their altitude i.e., to the lengths of the spans in which they 
occur ; thus the neg. mom.-areas (Fig. 446) GCH&n&DCH 
have a common base CH, 

, (M.A.y = r_ , 

* * (M.A.\" r r w 

(The notation explains itself ; see figure.) It also follows, 
that the resultant of these two neg. moment-areas (if re- 
quired in any construction ; see § 400) acts in a vertical 
which divides the horizontal distance between their gravity ver- 
ticals in the inverse ratio of the spans to which they belong 
[§ 21 and eq. (4) above]. 

Hence, since this horizontal distance is ^"+^"' their 
resultant must act in a vertical Y" r 9 whose distance from 
the gravity-vertical of GCH is %l"', and from that of 
CHI), %l. 

399. Amount and Gravity- Vertical of the Positive Moment 
Area of One Span as Due to Any Loading. — Since we can not 
deal directly with a continuous load by graphics, but must 
subdivide it into a number of detached loads sufficiently 
numerous to give a close approximation, let us suppose 
that this has already been done if necessary, and that P l9 
P 2 , etc., are the detached loads resting in the span AB in 
question ; see Fig. 447. 

Since [by (1), § 397] the positive moment-area is the 
same as the total moment-area would be if this portion of 
the beam simply rested on the extremities of the span, not 
extending beyond them, we may use the construction in § 
389 for finding it, remembering that in that paragraph the 



500 



MECHANICS OF ENGINEERING. 



oblique polygon in the lower part of Fig. 439 will serve 
as well as the (upper) one whose abutment -line is the 
beam itself, as far as moments are concerned. 

Hence, Fig. 447, lay off the load-line LL', take any pole 
0, with any convenient pole-distance H, and draw the 
equilibrium polygon FWG. After joining FG, FWGF 
will be the positive moment-area required. 

To find its gravity vertical, divide the span AB, or FG\ 
Into from ten to twenty equal parts (each =ds) and draw a 




Jte. 447. 



vertical through the middle of each. The lengths z lf % 
etc., on these verticals, intercepted in the moment-area, 
are proportional to the corresponding strips of moment- 
area, each of width =ds, and of an amount =HzAs. 

Form a load line, SK, of the successive s's, and with 
any pole 0', draw the equilibrium polygon A'B' (for the 
2-verticals). The intersection, B, of the extreme segments, 
is a point in the required gravity-vertical (§ 336). 

The amowit of the moment-area is (M.A.) 2 =Z[H'z4s'] 

=H.As. 2(z)=K48(z l +z s +z i + . . . ) 



CONTINUOUS GIRDER BY GRAPHICS. 501 

For example, with the span =£=120 in., subdivided into 
twelve equal As's, we have z/s=10 inches (of actual distance). 
If H—4l inches of paper and SK=I(z)=10.2 inches of paper, 
the force-scale being 80 lbs. to the inch, and the distance- 
scale 15 inches to the inch (1:15), we have 

(M.A. ) 2 = [4 x 80] x 10 x [10.2x15] =489600. (sq. in.)(lbs.) 

400. Construction of the "False-Polygons" For All the Spans 
of a Given (Prismatic) Continuous Girder, Under Given Loading, 
and With Given Heights of Supports. — [See note in § 397 for 
meaning of " false-polygon ". j Let us suppose that the 
given girder covers three unequal spans, Eig. 448, with 
supports at unequal heights, and that both extremities A 
and D are built-in, or " fixed," horizontally. To clear the 
ground for the present construction, we suppose that, from 
the given loading in each span, the positive moment-area 
of each span has been obtained in numerical form [so 
many (sq. in.) (lbs.) or (sq. in.) (tons)] and its gravity-ver- 
tical determined by § 398 or § 399 ; that the horizontal 
distances (i.e., the spans V, I", and V" and the distance be- 
tween the above gravity-verticals and the supports) have 
been laid off on some convenient scale ; that EI has been 
computed from the material and shape of section of the 
girder and expressed in the same units as the above mo- 
ment-areas ; that a convenient value for n has been se- 
lected (since EI-7-n is to be the pole distance of all the 
moment-area-diagrams), and that the vertical distances of 
B, C, and D, from the horizontal through A, have been 
laid off accordingly (see note in § 396). 

In the figure (448) verticals are drawn through the 
points of support ; also verticals dividing each span into 
thirds, since the unknown negative moment-areas (sub- 
scripts 1 and 3) act in the latter (§ 397) ; and the gravity- 
verticals of the known positive moment-areas. The ver- 
ticals Y' 9 T", V", and V"', are to be constructed later. 

The problem may now be stated as follows : 

Given the positions of the supports, the value of EI and n, 



502 



MECHANICS OF ENGINEERING. 



the fact that the girder is fixed horizontally at A and D, thi 
heights of supports, the loca tion of the gravity -verticals of all 
the positive and negative moment-areas, and the amounts of the 
'positive moment-areas ; it is required to find graphically the 
"false-polygon " in each span. 

The " false-polygons," viz.: A 123 B for the first span (on 
the left), B 123 C for the second, etc., are drawn in the figure 



<l i ■',/ &j \ ! I ("HI /i\' i T> 




i» ^4^J\i>v 










Fig. 448. 

for the purpose of discussing their properties at the out- 
set. Since SB and Bl are both tangent to the elastic curve 
at By they form a single straight line ; similarly (71 is but 
the prolongation of 3(7. Also Al and 3D must be hori- 
zontal since the beam is built in horizontally at its ex- 
tremities A and D. 

That is, the three false polygons form a continuous 
equilibrium polygon A . . . D, in equilibrium under the 
" loads " 

(KA.y, (M.A.y t (M.A.y, etc., 

so that we might use a single mom. -area-diagram in con- 
nection with it, but for convenience the latter will be 



COXTINrOUS GIRDER BY GRAPHICS. 503 

drawn in portions, one under each span, with a pole dis- 

, EI 
tance = 

n 

Of this polygon A ... D, we have the two segments Al 
and 3D already drawn, and know that it passes through 
the points B and G ; we shall next determine by construc- 
tion other points (called " fixed points ") p ', p", p Q ", p"', 
and Pq" in (the prolongations of) certain other segments. 

To find the "fixed point " p Q ' 7 where the segment 23 in the 
first span cuts the vertical V, the gravity -vertical of 
(M.A.\\ The vertex p' , or 1, is already known, being the 
intersection of Al with V. Lay off 2' 3' = (M.A.y which is 
known, and take a trial pole 0/with a pole-distance EI-^-n ; 
join 0/ 2' and 0/ 3'. Draw 12 || to 2'0 t to determine 2 
on the vertical (M. A.) 2 f , then through 2 a line || to 0/3' to 
cut V inp '. The unknown segment 23 must cut V : in 
the same point ; since all positions of 0/ on the vertical 
T'U' will result in placing p ' in this same point, and one 
of these positions must be the real pole 0' (unknown). 
[This is easily proved in detail by two pairs of similar 
triangles]. 

To determine the "fixed-point " p", in the prolongation 

of segment 12 of second span. The prolongations 

of the segments 12 (of first span) and 12 (of second span) 

must meet in a point k in the (vertical) line of action of the 

resultant of (M. A.y and (M. A.\" (§ 336). Although the 

amounts of (31. A.y and (M. A.)" are unknown, still the 

vertical line of action of their resultant (by § 398, Note) is 

I" 
known to be Y', a horizontal distance — - to the right of 

o 

(M.A.y ; hence Y' is easily drawn. Therefore, the unknown 
triangle &'31 has its three vertices on three known verti- 
cals, the side ¥3 passes through the known point p ', and 
the side 13 through the known point B. ' Now by pro- 
longing p ' 2 (or any line through p Q ') to cut (M. A.y and 
Y' in 3 and k f , respectively, joining 3 B and prolonging 
this line to cut (M. A.\" in some point 1 , and then joining 
&o' 1 , we have a triangle fto'3 lo °f which we can make a 



504 MECHANICS OF ENGINEEETNG. 

statement precisely the same as that just made for &' 3 1 

But if two triangles [as k'31 and & ' 3 1 ] have theii 
vertices on three parallel lines (or on three lines which 
meet in a point) the three intersections of their correspond- 
ing sides must lie on the same straight line [see reference 
to Chauvenet, § 378 a]. Of these intersections we have 
two,^ ' and B ; hence the third must lie at the intersection 
of the line p ' B (prolonged) with V 1 , and in this way the 
" fixed point" p", a point in lz'\ and ,\ in the segment 12 
(of second span; prolonged, is found. Draw a vertical 
through it and call it V". 

The fixed point p " (in prolongation of segment 23 of sec- 
ond span) lies in the vertical V" and is found from p" and 
the known value of {JSLA.\' f precisely as^V was found from 
p'. That is, we lay off vertically 2" 3" = (M.A.) a ", and join 
2" and 3" to t ", which is any point at distance EI — n 
to the right of 2"3". Through p" draw a line || to 2" t " 
to cut (31. A.) 2 " in 2 , then 2 p " II to t " 3" to determine 
pj' on the vertical V". 

The fixed points p'" andjV" i n ^ ne third span lie in the 
prolongations of the segments 12 and 23, respectively, of 
that span, p'" being found from the points p Q " and C and 
the verticals (H.A.)", Y", and (M.A.)"', in the same manner 
asp" was determined with similar data, while p '", in the 
same vertical V" as p"\ depends on (M. A.) 2 '" and its 
gravity vertical as already illustrated ; hence the detail 
need not be given ; see figure. 

In this way ior any number of spans we proceed from 
span to span toward the right and determine the succes- 
sive fixed points, until the points p and p of the last span 
have been constructed, which are p" and p f/ ' in our 
present problem. Since p '" is a point in the segment 23 
(prolonged) of the last span, we have only to join it with 
3 in that span, 'a point already known, and the segment 23 
is determined. Joining the intersection 2 with p'" we 
determine the next segment 21 and of coarse the vertex 1, 
which is then joined with G and prolonged to intersect 
M. A.)" to fix the segment 1(73 and the point 3. Join 



CONTINUOUS GIRDER BY GRAPHICS. 505 

3 p ", and proceed in a similar manner toward the left, 
until the whole equilibrium polygon (or series of " false 
polygons ") is finally constructed ; the last step being the 
joining of 2 with p'. 

401. Treatment of Special Features of the Last Problem. — (1.) 
If the beam is simply supported at A, Fig. 448, instead of 
built-in, (M. A.){ becomes zero, and the two segments A\ 
and 12 of that span form a single segment of unknown 
direction. Hence, the point A will take the place of p', 
and the vertical V A that of V. 

(2.) Similarly, if D, in the last span, is a simple support 
(beam not built in) (M. A.)$" becomes zero, and the seg- 
ments D3 and 32 form a single segment of unknown 
direction, so that after p /r has been found, we join p™ 
and D to determine the segment D2 ; i.e., in this last span, 
D takes the place of 3 of the previous article. 

(3.) If the first span carries no load (M. A.) 2 ' is zero, and 
the segments 12 and 23 will form a single segment 23. 
Hence if the beam is built in at A, p ' will coincide with 
the known pointy' (i.e., 1), while if A is a simple support 
p and p coincide with A, since, then, (M A.\' is zero and 
A 123 is a single segment. 

(4.) If the last span is unloaded (third span in Fig. 448), 
(M. A.) 2 '" is zero, 123 becomes a single segment, and hence 
Po'" will coincide with p"' ; so that after p p " has been con- 
structed it is to be directly joined to 3, if the beam is 
built in at D, and will thus determine the segment 13 ; or 
to D, if D is a simple support, (for then (M. A.)J" is zero 
and the three segments 12, 23, and 3 D form a single seg- 
ment.) 

(5.) If an intermediate span is unloaded (say the second 
span, Fig. 448) the positive mom.-area, (M. A.) 2 ", is zero, 
123 becomes a straight line, i.e. a single segment^ and 
therefore p" coincides with p" ; hence, when p" has been 
found we proceed as if it were p ". 

402. To Find the Negative Mom. -Areas, the Mom. -Curves, Shears, 
and Reactions of the Supports. — (1.) Having constructed the 



506 MECHANICS OF ENGINEERING. 

false polygons according to the last two articles, the nega* 
tive moment-areas of each span are then to be found by the 
note in § 397, Fig. 445, and expressed in numerical form. 

[If the positive mom.-area of the span is zero the points 
2' and 3' will coincide, Fig. 445, and in the case mentioned 
in (3), (or (4)), of § 401, if A (or D) were a simple support, 
in Fig. 448, the mom. -area-diagram of Fig. 445 would have 
but two rays.] 

(2.) The moments at the supports (or "end-moments " of 
the respective spans) depending, as they do, directly on 
the negative mom. -areas, can now be computed as illustrated 
in (3) § 397. The fact that each " end-moment " may be 
obtained from two negative mom.-areas, separately, one in 
each adjacent span (except, of course at the extremities of 
the girder) forms a check on the accuracy of the work. 
The two values should agree within one or two per cent. 

(3.) The " moment-curve " of each span or equilibrium 
polygon formed from a force-diagram whose load-line con- 
sists of the actual loads on the span laid off in proper 
order, can now be drawn, a convenient value for H having 
been selected (the same Hfor all the spans, that the moment- 
carves of successive spans may form a continuous line for 
the whole girder); since we may easily compute the proper 
moment ordinate at each support to represent the actual 
moment, then, for the H adopted, by (3) § 397. The 
moment-curve of each span, since we know its two extreme 
points and its pole-distance H, is then constructed by 
§ 341. 

(4.) The shear. Since the last construction involves 
drawing the special force-diagram for each span, with a 
ray corresponding to each part of the span between two 
consecutive loads, the shear at any section of the beam is 
easily found as being the length of the vertical projection 
of the " proper ray," interpreted by the force-scale of the 
force-diagram, as in §§ 389 and 390. With the shears as 



CONTINUOUS GIRDER BY GRAPHICS. 



507 




Fig. 449. 

iibrium, i.e., 2T= (§ 36), we have 



ordinates a shear -diagram may 
now be constructed, if desired, for 
each span. The directions of the 
shears should be carefully noted. 
(5.) Reactions of supports. Let 
us consider " free " the small por- 
tion of the girder, at each point 
of support, included between two 
sections, one close to the support 
on each side, Fig. 449. Suppose 
it is the support (7, and call the 
reaction, or pressure at that sup- 
port B c . Then, for vertical equi- 



ne — JcBr\~**c 



'CL 



(5) 



and, in general, the reaction at a support equals the 
(algebraic) sum of the two shears, one close to the support 
ou the right, the other on the left. The meaning of the 
subscripts is evident. In applying this rule, however, a 
free body like that in Fig. 449 should always be drawn, or 
conceived ; for the two shears are not always in the same 
direction ; hence the phrase " algebraic sum." 

At a terminal support, as A or F, Fig. 446, if the beam 
is not built in, the reaction is simply equal to the shear 
(since the beam does not overhang) just as in §§ 241 and 
243. Fig. 446 presents the peculiarity that the reaction 
of the support F is negative, (as compared with R in Fig. 
449); i.e., the support at F must be placed above the beam 
to prevent its rising (this might also be the case at C, or 
D, in Fig. 446, for certain relations between the loads). 



403. Numerical Example of Preceding Methods. — As illustrat- 
ing the constructions just given, it is required to investi- 
gate the case of a rolled wrought-iron " I-beam," [a 15- 
inch heavy beam of the N. J. Steel and Iron Co.,] extend- 
ing over four supports at the same level, covering three 



508 MECHANICS OF ENGINEERING. 

spans of 16, 20, and 14 feet respectively, and bearing a 
single load in each of the extreme spans, but a uniform 
load over the entire central span. As indicated thus : 

7 ft. 9 ft. 20 ft. 8 ft. 6 ft. 

A | B C | D 

p> w p»' 

30 tons 40 tons 32 tons 

[This is a practical case where W" is the weight of a 
brick wall, and P' and P"' are loads transmitted by col- 
umns from the upper floors of the building ; A and D are 
simple supports, and the weight of the girder is neglected.] 

The beam has a moment of inertia / = 707 biquad. 
inches, and the modulus of elasticity of the iron is E = 
25,000,000 lbs. per sq. in., = 12,500 tons per sq. in. 

Although with a prismatic continuous girder under 
given loading, with supports at the same level, it may easily 
be shown that the moments, shears, and reactions, to be 
obtained graphically, are the same for all values of 7, so 
long as the elastic limit is not surpassed, still, on account 
of the necessity of its use in other problems (supports 
not on a leveD. we shall proceed as if the value of I were 
essential in this one. 

Selecting the inch and ton as units for numerical work, 
we have 

EI = 12500 x 707. = 8,837,500 (sq. in.) (tons) while the 
respective positive mom.-areas, from eqs. (2) and (3) of 
§ 398, are : 

(M. A.y =% x 30 x 84 x 108 = 136,080 (sq. in.) (tons) 
(31. A.y = y i2 X 40 x 240 2 = 192,000 " " 
(M. A.) 2 '" = y 2 X 32 x 96 x 72 = 110,592 " " 

Adopting a scale of 60,000 (sq. in.) (tons) to the linear inch 
of paper, for mom. -area diagrams, we have for the above 
mom.-areas 2.27 in., 3.20 in., and 1.84 in., respectively, od 
the paper, for use in Fig. 448. 



CONTINUOUS GIRDER BY GRAPHICS. 509 

Having laid off the three spans on a scale of 60 inches 
to the inch of paper, with A, B, C, and D in the same hori- 
zontal line, we find by the construction of Fig. 443, that the 
gravity -vertical of (M. A.) 2 ' lies 3.6 in. to the left of the 
middle in the first span, that of (M. A.) 2 '" 4.8 in. to the 
right of the middle of the third span ; while that of 
(M. A.) 2 ", of course, bisects the central span. Hence we 
draw these verticals ; and also those of the unknown neg- 
ative mom.-areas through the one-third points ; remember- 
ing [§ 401, (1) and (2)] that (M. A.\' and (M. A.%'" are 
both zero in this case. 

Since EI = 8,837,500 (sq. in.) (tons), it would require 
147.29 in. to represent it, as pole-distance, on a scale of 
60,000 (sq. in.) (tons) to the inch ; hence let us take n = 50 
for the degree of (vertical) exaggeration of the false poly- 

gons, since the corresponding pole-distance — = 2.94 in. 

n 

of paper is a convenient length for use with the values of 
(M. A.) 2 ' t (M. A.y, etc., above given. 

Following the construction of Fig. 448, except that p' is 
at A, and p '" is to be joined to D (§ 401), (the student will 
do well to draft the problem for himself, using the pre- 
scribed scales,) and thus determining the false-polygons, 
we then construct and compute the neg. mom.-areas 
according to § 402 (1), and the note in § 397, obtaining the 
following results ; 

(M. A.\\ 1.43 in. of pap., = 85,800 (sq. in.) (tons) 
(M. A.\'\ 1.77 " " " = 106,200 " " 
(M. A.%", 1.68 " " " = 100,800 " " 
(M. A.\"\ 1.17" " " = 70,200 " " 



The remaining results are best indicated by the aid of 
Fig. 450. Following the items of § 402, we find [(3) § 397] 
that the moment at B, using (M. A.y, is 

M 2x106,200 = g85 inch . tons _ 
240 in. 



510 MECHANICS OF ENGINEERING. 

[or, using (31. A.) 3 ', \M* = 2x g^ 800 =893 in. tons.] 

Similarly, M c = 2xl00?80Q = 840 in. tons. 
[or, using (31 A.\'", 3I C = 835.7 in. tons.] 
Hence, taking means, we have, finally, 

3f x =0 ; if B =889 in. tons ; J_f c =837.8 ; if D =0. 

Fig. 450 shows the actual mom.-areas and shear-dia- 
grams, which are now to be constructed. 




14 ft-. 5J 



Fig. 450. 



Selecting a value IT = 20 tons for the pole-distance of 
the successive force-diagrams, (the scale of distances being 
5 ft. (60 in.) to the inch we have [(3) § 397] 

20 x ~BG = M B = 889in.-tons .-. BG = UA in. of actual 
distance, or 0.74 in. of paper; also 20 x GK = M G = 837.8 
in. -tons .-. GK = 41.89 in., or 0.698 in. of paper. 

Having thus found G and K, and divided BG into ten 
equal parts, applying four tons in the middle of each, we 
construct by § 341 an equilibrium polygon which shall 
pass through G and iTand have 20 tons as a pole-distance. 
(We take a force-scale of 10 tons to the inch.) It will 
form a (succession of short tangents to a) parabola, and is 
the moment curve for span BG. Similarly, for the single 



CONTINUOUS GIRDER BY GRAPHICS. 511 

loads P' and P'" in the other two spans, we draw the 
equilibrium polygons AN' G and KZ'"D, for the same H 
as before, and passing through A and G, and K and D, re- 
spectively. 

Scaling the moment -ordinates NN f , QQ", and ZZ'", 
reducing to actual distance and multiplying by H, we have 
for these local moment maxima, Jf N = 1008, M q = 336, and 
M z = 936, in. tons. 

Evidently the greatest moment is iJf N and .\ the stress 
in the outer fibre at N will be (§ 239) 

p y =j^= 1008x7 ^ = 10.0 tons per sq. incli which is much 

too large. If we employ a 20-inch heavy beam, with J = 
1650 biquad. in., the preceding moments will still be the 
same (supports all at same level) and we have 

1008x10 aria , 

_p x = — — — = 6.06 tons per sq. in., 

16o0 

or nearly 12,000 lbs. per sq. in., and is therefore safe 
(§183). 

If three discontinuous beams were to be used, the 20- 
inch size of beam (heavy) would be much too weak, in 
each of the three spans, as may be easily shown ; hence the 
economy of the continuous girder in such a case is readily per- 
ceived. It will be seen, however, that the cases of conti- 
nuity and of discontinuity do not differ so much in the 
shear-diagrams as in the moment curves. By scaling the 
vertical projection of the proper rays in the special force 
diagrams (as in §§ 389 and 390) we obtain the shear for 
any section on AN, as J XR (see Fig. 449 for notation) = 
12.3 tons ; on NB, J BL = 17.7 tons ; from B to G it varies 
uniformly from J BR = 20.3 tons, through zero at Q, to J Ch 
= 19.7 tons of opposite sign. Also, for CZ, J CR = 18.6 
tons ; and for ZD, J Dlj = 13.4 tons. Hence, the reactions 
of the supports are as follows : 

i? A =e/ AR =12.3 tons ; -5 B =«7 BL +«7 r BK =38.0 tons. 
B a =Jr jR -\- J" CL =38.3 tons ; B D =J Dll =13A tons. 



512 MECHANICS OF ENGINEERING. 

[In the shear-diagram, the shear-ordinates are laid off 
beloiv the axis when the shear points down, the "free body " 
extending to the right of the section considered, (as J CIj in Fig. 
449) ; and above, when the shear points upward for the 
same position of the free body.] 

If we divide the max. shear, 20.3 tons by the area of the 
web, 13.75 sq. in., of the 20-inch heavy beam, (§256), we 
obtain 1.5 tons or 3000 lbs. per sq. in., which is < 4000 
(§ 183). Notice the points of inflection, i', i 2 ", etc., where 
M is zero. 

Sufficient bearing surface should be provided at the 
supports. 

A swing-bridge offers an interesting case of a continu- 
ous girder, 

404. Continuous Girder of Variable Mom. of Inertia. — If i" is 
variable and I denote the mom. of inertia of some con- 
venient standard section, then we may write I = I -r- m, 
when m denotes the number of times I contains I. In a 
non-prismatic beam, m is different for different sections 
but is easily found, and will be considered given at each 
section. 

In eq. (1) of § 391, then, we must put I -f- m in place of 
i" and thus write 

d 2 y _[mMdx] ,~z 

M~ EI, ' : ' *. ' ' * ' W 

and (pursuing the same reasoning as there given) may 
therefore say that in a girder of variable section if each 
email vertical strip (Mdx) of the moment-area be multiplied by 
the value of m proper to that section, and these products (or "vir- 
tual mom. -area strips) considered as loads, the elastic curve is an 
equilibrium polygon for those loads with a pole distance = UI . 

In modifying § 400 for a girder of variable section, then,, 
besides taking EI Q -f- n as pole distance, proceed as 
follows : 

Construct the positive mom.-area for each span accord- 
ing to § 399 ; for each z of Fig. 447, substitute mz (each z 



COXTIXUOUS GIRDER BY GRAPHICS. 5i8 

having in general its own m), and thus obtain the " virtual 
positive mom.-area," and its gravity vertical. 

Similarly, there will be an unknown " virtual neg. mom.- 
area" not triangular, replacing each neg. mom.-area of 
§ 400. Though it is not triangular, each of its ordi- 
nates equals the corresponding ordinate of the unknown 
triangular neg. mom.-area multiplied by the proper m, and 
its gravity-vertical (which is independent of the amount of the 
unknown neg. mom.-area) is found in advance by the process 
of Fig. 447, using, for s's, a set of ordinates obtained thus : 
Draw any two straight lines AB and FB, Fig. 445, (for a 
left-hand trial neg. mom.-area; or FB and GFiov a right- 
hand one) meeting in the end-vertical of the span, divide 
the span into ten or twenty equal spaces, draw a vertical 
through the middle of each, noting their intercepts between 
AB and FB. Add these intercepts and call the sum S. 
Multiply each intercept by the proper m, and with these 
new values as «'s construct their gravity vertical as in Fig. 
447. Add these new intercepts, call the sum $ v , and denote 
the quotient S -f- S Y by /?. 

We substitute the three verticals mentioned, therefore, 
for the mom.-area verticals of § 400, and the " virtual pos. 
mom.-area " for the pos. mom.-area, in each span ; pro- 
ceed in other respects to construct the "false polygons" 
according to § 400. Then the result of applying the con- 
struction in the note § 397 will be the " virtual neg. mom.- 
areas," each of which is to be multiplied by the proper 
/3 to obtain the corresponding triangular neg. mom.-area, 
with which we then proceed, without further modifica- 
tions in the process, according to (2), (3), etc. of § 402. 

[The conception of these " virtual mom. -areas " is due 
to Prof. Eddy ; see p. 36 of his " Kesearches in Graphical 
Statics," referred to in the preface of this work.] 

405. Remarks. — It must be remembered that any unequal 
settling of the supports after the girder has been put in 
place, may cause considerable changes in the values of the 
moments, shears, etc., and thus cause the actual stresses to 
be quite different from those computed without taking 



5H XECHANICS OF ENGINEERING. 

into account a possible change in the heights of the sup- 
ports. See § 271. 

For example, if some of the supports are of masonry, 
while others are the upper extremities of high iron or 
steel columns, the fluctuations of length in the latter due 
to changes of temperature will produce results of the 
nature indicated above. 

If an open-work truss of homogeneous design from end 
to end (treated as a girder of constant moment of inertia, 
whose value may be formulated as in § 388,) is used as a 
continuous girder under moving loads, it will be subject 
to " reversal of stress " in some of its upper and lower hor- 
izontal members, i.e., the latter must be of a proper de- 
sign to sustain both tension and compression, (according 
to the position of the moving loads,) and this may disturb 
the assumption of homogeneity of design. Still, if i" is 
variable, § 404 can be used ; but since the weight of the 
truss must be considered as part of the loading, several 
assumptions and approximations may be necessary before 
establishing satisfactory dimensions. 



PART IV. 

HYDRAULICS. 



CHAPTER I. 
DEFINITIONS— FLUID PRESSURE-HYDROSTATICS BEGUN. 

406. A Perfect Fluid is a substance the particles of which 
are capable of moving upon each other with the greatest free 
dom, absolutely without friction, and are destitute of mutual 
attraction. In other words, the stress between any two con- 
tiguous portions of a perfect fluid is always one of compression 
and normal to the dividing surface at every point ; i.e., no 
shear or tangential action can exist on any imaginary cutting 
plane. 

Hence if a perfect fluid is contained in a vessel of rigid ma- 
terial the pressure experienced by the walls of the vessel is 
normal to the surface of contact at all points. 

For the practical purposes of Engineering, water, alcohol, 
mercury, air, steam, and all gases may be treated as perfect 
fluids within certain limits of temperature. 

407. Liquids and Gases. — A fluid a definite mass of which 
occupies a definite volume at a given temperature, and is in- 
capable both of expanding into a larger volume and of being 
compressed into a smaller volume at that temperature, is called 
a Liquid, of which water, mercury, etc., are common examples ; 
whereas a Gas is a fluid a mass of which is capable of almost 
indefinite expansion or compression, according as the space 
within the confining vessel is made larger or smaller, and al- 
ways tends to fill the vessel, which must therefore be closed in 
every direction to prevent its escape. 

515 



516 MECHANICS OF ENGINEEEING. 

Liquids are sometimes called inelastic fluids, and gases 
elastic fluids. 

408. Remarks. — Though practically we may treat all liquids 
as incompressible, experiment shows them to be compressible 
to a slight extent. Thus, a cubic inch of water under a pres- 
sure of 15 lbs. on each of its six faces loses only fifty millionths 
(0.000050) of its original volume, while remaining at the same 
temperature; if the temperature be sufficiently raised, how- 
ever, its bulk will remain unchanged (provided the initial tem- 
perature is over 40° Fahr.). Conversely, by heating a liquid in 
a rigid vessel completely filled by it, a great bursting pressure 
may be produced. The slight cohesion existing between the 
particles of most liquids is too insignificant to be considered in 
the present connection. 

The property of indefinite expansion, on the part of gases, 
by which a confined mass of gas can continue to fill a confined 
space which is progressively enlarging, and exert pressure 
against its walls, is satisfactorily explained by the " Kinetic 
Theory of Gases," according to which the gaseous particles are 
perfectly elastic and in continual motion, impinging against 
each other and the confining walls. Nevertheless, for prac- 
tical purposes, we may consider a gas as a continuous sub- 
stance. 

Although by the abstraction of heat, or the application of 
great pressure, or both, all known gases may be reduced to 
liquids (some being even solidified); and although by con- 
verse processes (imparting heat and diminishing the pressure) 
liquids may be transformed into gases, the range of tempera- 
ture and pressure in all problems to be considered in this work 
is supposed kept within such limits that no extreme changes of 
state, of this character, take place. A gas approaching the 
point of liquefaction is called a Vapor. 

Between the solid and the liquid state we find all grades of 
intermediate conditions of matter. For example, some sub- 
stances are described as soft and plastic solids, as soft putty, 
moist earth, pitch, fresh mortar, etc.; and others as viscous and 
sluggish liquids, as molasses and glycerine. In sufficient bulk, 



DEFINITION'S — FLUID PRESSURE — HYDROSTATICS. 517 

however, the latter may still be considered as perfect fluids. 
Even water is slightly viscous. 

409. Heaviness of Fluids. — The weight of a cubic unit of a 
homogeneous fluid will be called its heaviness, or rate of 
weight (see § 7), and is a measure of its density. Denoting it 
by y 9 and the volume of a definite portion of the fluid by V, 
we have, for the weight of that portion, 



G = Vy. 



(1) 



This, like the great majority of equations used or derived in 
this work, is of homogeneous form (§ 6), i.e., admits of any sys- 
tem of units. E.g., in the metre-kilogram-second system, if y 
is given in kilos, per cubic metre, V must be expressed in 
cubic metres, and G will be obtained in kilos.; and similarly 
in any other system. The quality of y, = G -i- V, is evidently 
one dimension of force divided by three dimensions of length. 

In the following table, in the case of gases, the temperature 
and pressure are mentioned at which they have the given 
heaviness, since under other conditions the heaviness would be 
different ; in the case of liquids, however, for ordinary pur- 
poses the effect of a change of temperature may be neglected 
(within certain limits). 



HEAVINESS OF VARIOUS FLUIDS.* 

[In ft. lb. sec. system; y — weight in lbs. of a cubic foot.] 



Liquids. 



Freshwater, y= 62.5 

Sea water 64.0 

Mercury 848.7 

Alcohol 49.3 

Crude Petroleum, about 55.0 

(N.B. — A cubic inch of water 
weighs 0.036024 lbs.; and a cubic 
foot 1000 av. oz.) 



fJn«p« J At tem P- of melting ice; and 14.7 
ua&tj&i lbs. per sq. in. tension. 



Atmospheric Air 0. 08076 

Oxygen 0.0892 

Nitrogen 0.0786 

Hydrogen 0. 0056 

Illuminating [from 0800 

Gas, J to 0.0400 

Natural Gas, about 0.0500 



* See Trautwine's Civ. Engineer's Pocket Book for an extended table — 
p. 380, edition of 1885. 



518 



MECHANICS OF ENGIIOJEIMN.G. 



For use in problems where needed, values for the heaviness 
of pure fresh water are given in the following table (from 
Rossetti) for temperatures ranging from freezing to boiling ; 
as also the relative density, that at the temperature of maxi- 
mum density, 39°. 3 Fahr. being taken as unity. The temper- 
atures are Fahr., and y is in lbs. per cubic foot. 



Temp. 


Rel. 
Dens. 


y- 


Temp. 


Rel. 

Dens. 


y.. 


Temp. 


Rel. 
Dens. 


V- 


32° 


.99987 


62.416 : 


60° 


.99907 


62.366 


140° 


.98338 


61.386 


35° 


.99996 


62.421, 


70° 


.99802 


62.300 


150° 


.98043 


61.203 


39°. 3 


1.00000 


62.424 


80° 


.99669 


62.217 


160° 


97729 


61.006 


40° 


.99999 


62.423' 


90° 


.99510 


62.118 


170° 


.97397 


60 799 


43° 


.99997 


62.422i 


100° 


.99318 


61.998 


180° 


.w056 


60.5b6 


45° 


.99992 


62.419 


110° 


.99105 


61.865 


190° 


.96701 


60.365 


50° 


.99975 


62.408! 


120° 


.98870 


61.719 


200° 


.96333 


60.135 


55° 


.99948 


62.390 


130° 


.98608 


61.555 


212° 


.95865 


59.843 



From D. K. Clark's] for temp. = 
"Manual." } y = 



230° 250° 270° 
59 4 58.7 58.2 


290° 

57.6 


298° 
57.3 


338° 
56.1 


366° 
55.3 



390° 
54.5 



Example 1. What is the heaviness of a gas, 432 cub. in. of 
which weigh 0.368 ounces? Use ft.-lb.-sec. system. 

432 cub. in. = \ cub. ft. and 0.368 oz. = 0.023 lbs. 



,\ y = - = -'— — = 0.092 lbs. per cub. foot. 



Example 2. Required the weight of a right prism of mer- 
cury of 1 sq. inch section and 30 inches altitude. 

30 
1^=30 X 1 = 30 cub. in. = — — - cub. feet : while from the 

1728 

table, y for mercury = 848.7 lbs. per cub. ft. 

.\ its weight = G = Vy = -^-- X 848.7 = 14.73 lbs. 

1 /28 



410. Definitions. — By Hydraulics we understand the me- 
chanics of fluids as utilized in Engineering. It may be divided 
into 

Hydrostatics, treating of fluids at rest ; and 

Hydrodynamics (or Hydrokinetics), which deals with fluids 
in motion. (The name Pneumatics is sometimes used to cover 
both the statics and dynamics of gaseous fluids.) 



DEFINITIONS — FLUID PRESSURE — HYDROSTATICS. 519 

[Rankine's nomenclature has been adopted in the present 
work. Some recent writers use the term Hydromechanics for 
mechanics of fluids, subdividing it into Hydrostatics and 
HydroMnetics, as above ; they also use the term Dynamics to 
embrace both of the two divisions called Statics and Dynamics 
by Rankine, which by them are called Statics and Kinetics re- 
spectively. Though unusual, perhaps, the term Hydraulics is 
here used to cover the applied Mechanics of Gases as well as 
of Liquids.] 

Before treating separately of liquids and gases, a few para- 
graphs will be presented applicable to both kinds of fluids. 

411. Pressure per Unit Area, or Intensity of Pressure. — As in 
§ 180 in dealing with solids, so here with fluids we indicate the 
pressure per unit area between two contiguous portions of 
fluid, or between a fluid and the wall of the containing vessel, 
by p, so that if dP is the total pressure on a small area dF, 
we have 

dP rts 

* = W (1 > 

as the pressure per unit area, or intensity of pressure (often, 
though ambiguously, called the tension in speaking of a gas) 
on the small surface dF. If pressure of the same intensity 
exists over a finite plane surface of area = F, the total pres- 
sure on that surface is 

P = fpdF=pfdF= Fp, "1 

P [ .... (2) 

or #=y. J 

(N.B. — For brevity the single word " pressure" will some- 
times be used, instead of intensity of pressure, where no am- 
biguity can arise.) Thus, it is found that, under ordinary con- 
ditions at the sea level, the atmosphere exerts a normal pressure 
(normal, because fluid pressure) on all surfaces, of an intensity 
of about p = 14.7 lbs. per sq. inch (= 2116. lbs. per sq. ft.). 
This intensity of pressure is called one atmosphere. For ex- 



520 



MECHANICS OF ENGINEERING. 



ample, the total atmospheric pressure on a surface of 100 sq, 
in. is [inch, lb., sec] 

P = Fp = 100 X 14.7 = 1470 lbs. (.= 0.735 tons.) 

The quality of p is evidently one dimension of force divid- 
ed by two dimensions of length. 

412. Hydrostatic Pressure; per Unit Area, in the Interior of a 
Fluid at Rest. — In a body of fluid of uniform heaviness, at 
rest, it is required to find the mutual pressure per unit area be- 
tween the portions of fluid on opposite sides of any imaginary 
cutting plane. As customary, we shall consider portions of 
the fluid as free bodies, by supplying the forces' exerted on 
them by all contiguous portions (of fluid or vessel wall), also 
those of the earth (their weights), and then apply the condi- 
tions of equilibrium. 

First, cutting plane horizontal. — Fig. 451 shows a body of 
homogeneous fluid confined in a rigid 
vessel closed at the top with a small air- 
tight but frictionless piston (a horizontal 
disk) of weight = G and exposed to at- 
mospheric pressure (=p a P er unit area ) 
on its upper face. Let the area of piston- 
face be = J?> Then for the equilibrium 
of the piston the total pressure between 
its under surface and the fluid at O must 
be 

I>=G + F Pa , 




Fig. 451. 



and hence the intensity of this pressure is 



G 



Po = -^+Pa- 



(1) 



It is now required to find the intensity,^?, of fluid pressure 
between the portions of fluid contiguous to the horizontal cut- 
ting plane i^at a vertical distance = h vertically below the pis- 
ton O. In Fig. 452 we have as a free body the right parallelo- 



FLUID PKESSTTKE. 



521 



piped OBC of Fig. 451 with vertical sides (two || to paper and 
four "1 to it). The pressures acting on its six faces are normal 
to them respectively, and the weight of the prism is = vol. 
Xy = Fhy, supposing y to have the same value at all parts of 
the column (which is practically true for any height of liquid 
and for a small height of gas). Since the , — >~~^ v 

prism is in equilibrium under the forces ollilj 

shown in the figure, and would still be so 
were it to become rigid, we may put (§ 36) 
2 (vert, compons.) == and hence obtain 

F#-Fp - Fhy = 0. . . (2) 

(In the figure the pressures on the ver- 
tical faces || to paper have no vertical com- 
ponents, and hence are not drawn.) From fig. 452. 
(2) we have 



>-! 

>\ 


|- 








D 


c 




3P 



P =i>. + hy. 



(3) 



(hy, being the weight of a column of homogeneous fluid of unity 
cross-section and height h. would be the total pressure on the 
base of such a column, if at rest and with no pressure on the 
upper base, and hence might be called intensity due to weight.) 
Secondly, cutting plane oblique. — Fig. 453. Consider free 
an infinitely small right triangular prism bed, whose bases are 

|| to the paper, while the three side 
faces (rectangles), having areas = dF, 
dF, , and dF 3 , are respectively hori- 
zontal, vertical, and oblique ; let angle 
cbd = a. The surface be is a portion 
— W Fl of the plane BG of Fig. 452. Given 
- — \ P( = intensity of pressure on dF) and 
J a, required^, the intensity of pressure 
on the oblique face bd, of area dF n . 
[N. B. — The prism is taken very small 
in order that the intensity of pressure may be considered con- 
stant over any one face ; and also that the weight of the prism 
may be neglected, since it involves the volume (three dimen- 




FlG. 453. 



522 MECHANICS OF ENGINEERING. 

sions) of the prism, while the total face pressures involve only 
two, and is hence a differential of a higher order.] 
From 3 (vert comports.) = 0we shall have 

p t dF 2 cos a — pdF = ; but dF -± dF 2 = cos a ; 

1>*=1>> W 

which is independent of the angle a. 

Hence, the intensity of fluid pressure at a given point is 
the same on all imaginary cutting planes containing the 
point. This is the most important property of a fluid, and is 
true whether the liquid is at rest or has any kind of motion ; 
for, in case of rectilinear accelerated motion, e.g., although the 
sum of the force-components in the direction of the accelera- 
tion does not in general = 0, but = mass X ace, still, the 
mass of the body in question is = weight -~ g, and therefore 
the term mass X ace. is a differential of a higher order than 
the other terms of the equation, and hence the same result 
follows as when there is no motion (or uniform rectilinear 
motion). 

413. The Intensity of Pressure is Equal at all Points of any 
Horizontal Plane in a body of homogeneous fluid at rest. If 
we consider a right prism of the fluid in Fig. 451, of small 
vertical thickness, its axis lying in any horizontal plane BC, 
its bases will be vertical and of equal area dF. The pressures 
on its sides, being normal to them, and hence to the axis, have 
no components || to the axis. The weight of the prism also 
has no horizontal component. Hence from 2 (nor. comps. 
|| to axis) = 0, we have, p x and p 3 being the pressure-intensi- 
ties at the two bases, 

p 1 dF—p z dF=0; .'.p=p 3 , . . . . (1) 

which proves the statement at the head of this article. 

It is now plain, from this and the preceding article, that 
the pressure-intensity p at any point in a homogeneous fluid 
at rest is equal to that at any higher point, plus the iveight 



FLUID PRESSURE. 



523 



(hy) of a column of the fluid of section unity and of altitude 
(A) = vertical distance betiveen the points. 



p=p* + h r> 



(2> 



whether they are in the same vertical or not, and whatever be 
the shape of the containing ^ 
{or pipes), provided the 



S 



fluid is continuous between 
the two points; for, Fig. 454, 
by considering a series of 
small prisms, alternately ver- 
tical and horizontal, obcde, we 
know that 




==±-ft.o 



Fig.454. 



Pe =Pb ; 

and p e =jp d ; 



Pb=P* + fal 

Pd=Po — Kr\ 

hence, finally, by addition we have 

Pe = 'p + hy 

(in which h = A, — h 2 ). 

If, therefore, upon a small piston at o, of area = F , a force 
P be exerted, and an inelastic fluid (liquid) completely fills the 
vessel, then, for equilibrium, the force to be exerted upon the pis- 
ton at 0, viz., P e , is thus computed : For equilibrium of fluid 
p e =p -f. hy ; and for equil. of piston o, p — P -f- F ; also, 



Pe 



F - 



:P. 



F e hy 



(3) 



From (3) we learn that if the pistons are at the same level 
(h, = 0) the total pressures on their inner faces are directly 
proportional to their areas. 

If the fluid is gaseous (2) and (3) are practically correct if 
h is not > 100 feet (for, gas being compressible, the lower 
strata are generally more dense than the upper), but in (3) the 
pistons must be fixed, and P e and P refer solely to the in- 
terior pressures. 



524 MECHANICS OF ENGINEERING. 

Again, if h is small or p very great, the terra hy may be 
omitted altogether in eqs. (2) and (3) (especially with gases, 
since for them y (heaviness) is usually small), and we then 
have, from (2), 

jp=i>.; (4) 

being the algebraic form of the statement : A body of fluid 
at rest transmits pressure with equal intensity in every direc- 
tion and to all of its parts. [Principle of "Equal Transmis- 
sion of Pressure."] 

414. Moving Pistons. — If the fluid in Fig. 454 is inelastic 
and the vessel walls rigid, the motion of one piston (p) through 
a distance s Q causes the other to move through a distance s e de- 
termined by the relation F Q s = F e s e (since the volumes de- 
scribed by them must be equal, as liquids are incompressible) ; 
but on account of the inertia of the liquid, and friction on the 
vessel walls, equations (2) and (3) no longer hold exactly, still 
are approximately true if the motion is very slow and the 
vessel short, as with the cylinder of a water-pressure engine. 

But if the fluid is compressible and elastic (gases and vapors ; 
steam, or air) and hence of small density, the effect of inertia 
and friction is not appreciable in short wide vessels like the 
cylinders of steam- and air-engines, and those of air-compres- 
sors ; and eqs. (2) and (3) still hold, practically, even with high 
^^ piston-speeds. For example, in the space AB, 

Fig. 455, between the piston and cylinder-head 
f of a steam-engine (piston moving toward the 



j — 2 right) the intensity of pressure, p, of the 

-\ steam against the moving piston B is prac- 



Fm. 455. tically equal to that against the cylinder-head 

A at the same instant. 

415. An Important Distinction between gases and liquids 
(i.e., between elastic and inelastic fluids) consists in this : 

A liquid can exert pressure against the walls of the contain- 
ing vessel only by its weight, or (when confined on all sides) 
by transmitted pressure coming from without (due to piston 
pressure, atmospheric pressure, etc.) ; whereas — 



FLUID PRESSURE. 525 

A gas, confined, as it must be, on all sides to prevent dif- 
fusion, exerts pressure on the vessel not only by its weight, 
but by its elasticity or tendency to expand. If pressure from 
without is also applied, the gas is compressed and exerts a still 
greater pressure on the vessel walls. 

416. Component, of Pressure, in a Given Direction. — Let 
ABCD, whose area = dF, be a small element of a surface, 
plane or curved, and^> the intensity of a 

fluid pressure upon this element, then ^ F \ /iX 
the total pressure upon it is pdF, and is N/ \ A '\ B 

of course normal to it. Let A'B' CD be /^'^y\ 
the projection of the element dF upon cr"" / \\b^'"^ 

a plane CDM making an angle a with V* / ^^' 
the element, and let it be required to j 

find the value of the component of pdF FlG * 455 * 

in a direction normal to this last plane (the other component 
being understood to be || to the same plane). We shall have 

Compon. of pdF ~] to CDM = pdF cos a — p{dF . cos a). (1) 

But dF . cos a = area A'B'CD, the projection of dFw^on 
the plane CDM. 

.: Compon. ~\ to plane CDM =p X {project, of dF on CDM); 

i.e., the component of fluid pressure (on an element of a sur- 
face) in a given direction (the other component being "| to 
the first) is found by multiplying the intensity of the pressure 
by the area of the projection of the element upon a plane ~\ to 
the given direction. 

It is seen, as an example of this, that if the fluid pressures 
on the elements of the inner surface of one hemisphere of a 
hollow sphere containing a gas are resolved into components "1 
and || to the plane of the circular base of the hemisphere, the 
sum of the former components simply = 7tr 2 p, where r is the 
radius of the sphere, and^j> the intensity of the fluid pressure ; 
for, from the foregoing, the sum of these components is just 
the same as the total pressure would be, having an intensity p. 



626 



MECHANICS OF ENGINEEKING. 



oil a great circle of the sphere, the area, tt/' 2 , of this circle being 
the sum of the areas of the projections, upon this circle as a 
base, of all the elements of the hemispherical surface. (Weight 
of fluid neglected.) 

A similar statement may be made as to the pressures on 
the inner curved surface of a right cylinder. 

417. Non-planar Pistons. — From the foregoing it follows that 
the sum of the components || to the piston-rod, of the fluid 
pressures upon the piston at A, Fig. 457, is just the same as at 
jB, if the cylinders are of equal size and the steam, or air, is at 
the same tension. For the sum of the projections of all the 
elements of the curved surface of A upon a plane "1 to the 
piston-rod is always = nr* = area of section of cylinder-bore. 




£ 



Fig. 457. 



If the surface of A is symmetrical about the axis of the cylin- 
der the other components (i.e., those "1 to the piston-rod) will 
neutralize each other. If the line of intersection of that sur- 
face with the surface of the cylinder is not symmetrical about 
the axis of the cylinder, the piston may be pressed laterally 
against the cylinder-wall, but the thrust along the rod or 
" working force 7 (§ 128) is the same (except for friction in- 
duced by the lateral pressure), in all instances, as if the surface 
were plane and 1 to piston-rod. 

418. Bramah, or Hydraulic, Press. — This is a familiar instance 
of the principle of transmission of fluid pressure. Fig. 458, 
Let the small piston at have a diameter d = 1 inch = -fa ft., 
while the plunger E, or large piston, has a diameter d' — AB 
— CD =16 in. == f ft. The lever MJV weighs G, = 3 lbs., 
and a weight G = 40 lbs. is hung at M. The lever-arms of 
these forces about the fulcrum W are given in the figure. 
The apparatus being full of water (oil is often used), the fluid 
pressure P against the small piston is found by putting 



FLUID PRESSURE. 



527 



2(moms. about If) = for the equilibrium of the lever; 
whence [ft., lb., sec] 

P o x 1 _ 40 X 3 - 3 X 2 = 0. .-. P = 126 lbs. 



j, 2 - 



Gi 



£ 



4_»j 

3b 



HI 



Wc 



U E 



A B 



C e D 



Fig. 458. 

But, denoting atmospheric pressure by p a , and that of the 
water against the piston by p (per unit area), we may also 
write 

-Po = ^>o - F»Pa = i nd\<p -p a ). 

Solving for p , we have, putting p a = 14.7 X 144 lbs. per 
sq. ft., 

p = [l26 -5- 1 (^) 2 1 + 14.7 X1M = 25236 lbs. per sq. ft. 

Hence at e the press, per unit area, from § 409, and (2), § 413, is 
p e = p o -f hy = 25236 + 3 X 62.5 = 25423 lbs. per sq. ft. 

= 175.6 lbs. per sq. inch or 11.9 atmospheres, and the total 
upward pressure at e on base of plunger is 

P = F e p e = n ^p e = i 7t(j.y X 25423 = 31194 lbs., 

or almost 16 tons (of 2000 lbs. each). The compressive force 
upon the block or bale, C, = P less the weight of the plunger 
and total atmos. pressure on a circle of 15 in. diameter. 



528 



MECHANICS OF ENGINEERING. 



41 9, The Dividing Surface of Two Fluids (which do not mix) in 
Contact, and at Rest, is a Horizontal Plane. — For, Fig. 459, sup- 
posing any two points e and of this sur- 
face to be at different levels (the pressure 
at being p , that at ep e , and the heavi- 
nesses of the two fluids y x and y a respec- 
tively), we would have, from a considera- 
tion of the two elementary prisms eb and 
h (vertical and horizontal;, the relation 




Fig. 459. 



2>e=p a + hy y \ 
while from the prisms eo and cO, the relation 

Pe =P + h Y, • 

These equations are conflicting, hence the aoove supposition 
is absurd. Therefore the proposition is true. 

For stable equilibrium, evidently, the heavier fluid must oc- 
cupy the lowest position in the vessel, and if there are several 
fluids (which do not mix), they will arrange 
themselves vertically, in the order of their den- 
sities, the heaviest at the bottom, Fig. 460. On 
account of the property called diffusion the par- 
ticles of two gases placed in contact soon inter- 
mingle and form a uniform mixture. This fact 
gives strong support to the " Kinetic Theory of 
Gases" (§ 408). 




Fig. 460. 



420. Free Surface of a Liquid at Rest. — The surface (of a 
liquid) not in contact with the walls of the containing vessel 
..._,;..._...,.-,;,.:-, is called a free surface, and is necessarily 
horizontal (from § 419) when t .e liquid is at 
rest. Fig. 461. (A gas, from its tendency 
to indefinite expansion, is incapable of hav- 
ing a free surface.) This is true even if the 
space above the liquid is vacuous, for if the 
ft surface were inclined or curved, points in the 
body of the liquid and in the same horizon- 
tal plane would have different heights (or " heads") of liquid 





T .:■-'-:■ :v'.V''fl 


'•'! ! 


•;':' , 'v;.'-~AlR-" : .':'.' ; '^' 


i ; : 


( '• "•' • ' 




:-~i:y:-j>;^--:::-.j r 




— 














— 




Z^WATER— -^T 


T,/. 


Y///////////////////////////////} 



Fig. 461. 



TWO LIQUIDS IN BENT TUBE. 



529 



between them and the surface, producing different intensities 
of pressure in the plane, which is contrary to § 413. 

When large bodies of liquid like the ocean are considered, 
gravity can no longer be regarded as acting in parallel lines ; 
consequently the free surface of the liquid is curved, being ~] 
to the direction of (apparent) gravity at all points. For ordi- 
nary engineering purposes (except in Geodesy) the free surface 
of water at rest is a horizontal plane. 

421. Two Liquids (which do not mix) at Rest in a Bent Tube 
open at Both Ends to the Air, Fig. 460 ; water and mercury, for 
instance. Let their heavinesses be y L ..->>-, 

and y^ respectively. The pressure at e 
may be written (§ 413) either 



or 



■■Po, + Ky x 

p% + hy* 



according as we refer it to the water 
column or the mercury column and 
their respective free surfaces where the 
pressure^ =Po 2 = Pa = atmos. press. 
e is the surface of contact of the two liquids. 




Fig. 462. 



Hence we have 



Pa + Kr< =p a + Kr% ; i.e., K '• Km y* 



(3) 



i.e., the heights of the free surfaces of the two liquids above the 
surface of contact are inversely proportional to their respec- 
tive heavinesses. 

Example. — If the pressure at e = 2 atmospheres (§ 396) we 
shall hav^ from (1) (inch-lb.-sec. system of units) 

KY* =&6—Pa = 2X 14.7 — 14.7 = 14.7 lbs. per sq. inch. 
/. h u must = 14.7 -r- [848.7 -f- 1728] = 30 inches 

(since, for mercury, y 2 = 848.7 lbs. per cub. ft.). Hence, 
from (3), 

, \y % 30 X [848.7 ~- 1728] , _ Q . , 

K = — = ^ 5. + 1W8 = 408 ^hes - 34 feet. 



530 MECHANICS OF ENGINEERING. 

i.e., for equilibrium, and that j? e may == 2 atmospheres, A 3 aud 
A a (of mercury and water) must be 30 in. and 34 feet respec- 
tively. 

422. City Water-pipes. — If h = vertical distance of a point 
B of a water-pipe below the free surface of reservoir, and the 
water be at rest, the pressure on the inner surface of the pipe 
at B (per unit of area) is 

p = <p o -{- ky ; and here^> =p a = atmos. press. 

Example. — If h — 200 ft. (using the inch, lb. ? and second) 

p = 14.7 + [200 X 12] [62.5 -£- 1728] = 101.5 lbs. per sq. in. 

The term hy. alone, = 86.8 lbs. per sq. inch, is spoken of as the 
hydrostatic pressure due to 200 feet height, or "Head," of 
water. (See Trautwine's Pocket Book for a table of hydro- 
static pressures for various depths.) 

If, however, the water \& flowing through the pipe, the pres- 
sure against the interior wall becomes less (a problem of Hy- 
drodynamics to be treated subsequently), while if that motion 
is suddenly checked, the pressure becomes momentarily much 
greater than the hydrostatic. This shock is called '• water- 
ram" and " water-hammer," and may be as great as 200 to 300 
lbs. per sq. inch. 

423. Barometers and Manometers for Fluid Pressure. — If a 
tube, closed at one end, is filled with water, and the other ex- 
tremity is temporarily stopped and afterwards 
opened under water, the closed end being then 
a (vertical) height = h above the surface of 
the water, it is required to find the intensity, 
p Q , of fluid pressure at the top of the tube, sup- 
posing it to remain filled with water. Fig. 
463. At E inside the tube the pressure is 
14.7 lbs. per sq. inch, the same as that outside 
at the same level (§ 413) ; hence, from p E =j> a 

+ hy, 

Po=PE-by (I)' 




BAROMETERS. 531 

Example. — Let h = 10 feet (with inch-lb.-sec. system) ; then 
p = 14.7 - 120 X [62.5 ~ 1728] = 10.4 lbs. per sq. inch, 

or about f of an atmosphere. If now we inquire the value 
of h to make j? = zero, we put^> £ — hy = and obtain h = 
40S inches, = 34 ft., which is called the height of the water- 
barometer. Hence, Fig. 4:63a, ordinary atmospheric pressure 
will not sustain a column of water higher than 34 feet. If 
mercury is used instead of water the height supported by one 
atmosphere is 

b =5 14.7 ~ [848.7 -f- 1728] = 30 inches, 



$ 



Fig. 463a. 



= 76 centims. (about), and the tube is of more manageable 
proportions than with water, aside from the ad- 
vantage that no vapor of mercury forms above 
the liquid at ordinary temperatures. [In fact, the 
water-barometer height b = 34 feet has only a 
theoretical existence since at ordinary tempera- 
tures (40° to 80° Fahr.) vapor of water would 
form above the column and depress it by from 
0.30 to 1.09 ft.] Such an apparatus is called a 
Barometer, and is used not only for measuring 
the varying tension of the atmosphere (from 14.5 
to 15 lbs. per sq. inch, according to the weather and height 
above sea-level), but also that of any body of gas. Thus, Fig. 
464, the gas in D is put in communication with 
the space above the mercury in the cistern at 
C; and we have p = hy, where y = heav. of 
mercury, and p is the pressure on the liquid in 
the cistern. For delicate measurements an at- 
tached thermometer is also used, as the heavi- 
ness y varies slightly with the temperature. 

If the vertical distance CD is small, the ten- 
sion in C is considered the same as in D. 

For gas-tensions greater than one atmosphere, 
the tube may be left open at the top, forming an open ma- 



^■p-\ 



Fig. 464. 



532 



MECHANICS OF ENGINEEKING. 



nometer, Fig. 465. In this case, the tension of the gas above 
the mercury in the cistern is 



j> = (h + b)y, 



(1) 



in which b is the height of mercury (about 30 
pyil in.) to which the tension of the atmosphere above 
the mercury column is equivalent. 



Fig. 465. 



Example. — If h — 51 inches, Fig. 465, we 
have (ft,, lb., sec.) 



p = [4.25 ft. + 2.5 ft.] 848.7 = 5728 lbs. per sq. foot 
= 39.7 lbs. per sq. incli = 2.7 atmospheres. 



Another form of the open manometer consists of a U tube, 
Fig. 464, the atmosphere having access to one branch, the gas 
to be examined, to the other, while the - r;;> , 

mercury lies in the curve. As before, we ,vM^M : i^. 

have ;';^,^^ ;V 



p = (h + b)y = hy +p c 



(2) 



«S*1 



c 



where p a = atmos. tension, and b as above. 
The tension of a gas is sometimes spoken 
of as measured by so many inches of mer- ^v^-v-^;] ^^0 
cury. For example, a tension of 22.05 fig. 466. 

lbs. per sq. inch (1J- atmos.) is measured by 45 inches of mer- 
cury in a vacuum manometer (i.e., a common barometer), 
Fig. 464. "With the open manometer this tension (1-J atmos.) 
would be indicated by 15 inches of actual mercury, Figs. 465 
and 466. An ordinary steam-gauge indicates the excess of 
tension over one atmosphere ; thus " 40 lbs. of steam" implies 
a tension of 40 -f- 14.7 = 54.7 lbs. per sq. in. 

The Bourdon steam-gauge in common use consists of a 
curved elastic metal tube of flattened or elliptical section 
(with the long axis ~| to the plane of the tube), and has one 
end fixed. The movement of the other end, which is free and 



TENSION OF GASES. 



533 



closed, by proper mechanical connection gives motion to the 
pointer of a dial. This movement is caused by any change of 
tension in the steam or gas admitted, through the fixed end, to 
the interior of the tube. As the tension increases the ellip- 
tical section becomes less flat, i.e., more nearly circular, caus- 
ing the two ends of the tube to separate more widely, i.e., the 
free end moves away from the fixed end ; and vice versa. 

Such gauges, however, are not always reliable. They are 
graduated by comparison with mercury manometers ; and 
should be tested from time to time in the same way.* 

424. Tension of Illuminating Gas. — This is often spoken of as 
measured by inches of water (from 1 to 3 inches usually). 
Strictly it should be stated that this 
water-height measures the excess of 
its tension over that of the atmos- 
phere. Thus, in Fig. 466, water 
being used instead of mercury, h = 
say 2 inches, while b = 408 inches. 

This difference of tension may be 
largely affected by a change in the 
barometer due to the weather, or by 
a difference in altitude, as the follow- 
ing example will illustrate : 

Example. — Supposing the gas at rest, and the tension at the 
gasometer A, Fig. 467, to be " two inches of water," required 
the water-column Ji" (in open tube) that the gas will support 
in the pipe at B, 120 feet (vertically) above the gasometer. 
Let the temperature be freezing (nearly), and the outside air at 
a tension of 14.7 lbs. per sq. inch ; the heaviness of the gas at 
this temperature being 0.036 lbs. per cubic foot. For the 
small difference of 120 ft. we may treat both the atmosphere 
and the gas as liquids, that is, of constant density throughout 
the vertical column, and therefore apply the principles of 
§ 413 ; with the following result : 

The tension of the outside air at _B, supposed to be at the 
same temperature as at A, will sustain a water-column less 
than the 408 inches at A by an amount corresponding to the 

* Of late years gauges have come into use constructed of boxes with cor- 
rugated sides of thin metal like the aneroid barometer. Motion of the sides, 
under varying internal fluid pressure, causes movement of a pointer on a dial. 




Fig. 467. 



534 MECHANICS OF ENGINEEEING. 

120 feet of air between, of the heaviness .0807 lbs. per cub. 
ft. 120 feet of air weighing .0807 lbs. per cub. ft. will balance 
0.154 ft. of water weighing 62.5 lbs. per cubic ft., i.e., 1.85 
inches of water. ~Now the tension of the gas at B is also less 
than its tension at A, but the difference is not so great as with 
the outside air, for the 120 ft. of gas is lighter than the 120 ft. 
of air. Since 120 ft. of gas weighing 0.036 lbs. per cubic ft. 
will balance 0.0691 ft., or 0.83 inches, of water, therefore the 
difference between the tensions of the two fluids at B is greater 
than at A by (1.85 - 0.83 = ) 1.02 inches ; or, at B the total 
difference is 2.00 -f 1.02 = 3.02 inches. 

Hence if a small aperture is made in the pipe at B the gas 
will flow out with greater velocity than at A. At Ithaca, 
N. Y., where the University buildings are 400 ft. above the 
gas-works, this phenomenon is very marked. 

When the difference of level is great the decrease of tension 
as we proceed upward in the atmosphere, even with constant 
temperature, does not follow the simple law of § 413 ; see 
§477. 

For velocity of flow of gases through orifices, see § 548, etc. 

425. Safety-valves. — Fig. 468. Required the proper weight 

G to be hung at the extremity of the horizontal lever AB, 

with fulcrum at B, that the flat 

disk- valve E shall not be forced 

upward by the steam pressure, £>', 

until the latter reaches a given 

' ■^ ; .-gv:.'^-.-::-^%ga^ value =p. Let the weight of 

~-~w ■•■ ' .: U- ^ ^he arm i^ q j^-g cen tre of grav- 

Fig. 468. .- _ . *' .. & 

lty being at 6, a distance = o 

from B ; the other horizontal distances are marked in the 

figure. 

Suppose the valve on the point of rising ; then the forces 

acting on the lever are the fulcrum-reaction at B, the weights 

G and G 1 , and the two fluid-pressures on the disk, viz. : Ep a 

(atmospheric) downward, and Ep (steam) upward. Hence, 

from 2(moms. B ) = 0, 

Gb + G,c + Ep a a - Epa = 0. . . . (1) 




BURSTING OF PIPES. 535 

Solving, we have 

G^Fip-^-Gt. (2) 

Example. — With a = 2 inches, b — 2 feet, c = 1 foot 
G 1 = 4 lbs.," j? = 6 atmos., and diam. of disk = 1 inch ; with 
the foot and pound, 

G = L'l (f2) 2[6 X U ' 7 X 1M - X X 14 ' 7 x 144 ] - 4 X*. 
.-. G = 2.81 lbs. 

[Notice the cancelling of the 144; for F{p — p a ) i& pounds, 
being one dimension of force, if the pound is selected as the 
unit of force, whether the inch or foot is used in both fac- 
tors.] Hence when the steam pressure has risen to 6 atmos. 
(= 88.2 lbs. per square inch) (corresponding to 73.5 lbs. per sq. 
in. by steam-gauge) the valve will open if G = 2.81 lbs., or be 
on the point of opening. 

426. Proper Thickness of Thin Hollow Cylinders (i.e., Pipes 
and Tubes) to Resist Bursting by Fluid Pressure. 

Case I. Stresses in the cross-section due to End Pressure* 
Fig. 469. — Let AB be the circular cap clos- 
ing the end of a cylindrical tube containing 
fluid at a tension = p. Let r — internal 
radius of the tube or pipe. Then considering 
the cap free, neglecting its weight, we 
have three sets of || forces in equilibrium 
in the figure, viz. : the internal fluid pres- 
sure = nr 2 p ; the external fluid pressure 
= nr*p a ; while the total stress (tensile) on 
the small ring, whose area now exposed is 
27trt (nearly), is = %7trtp 1 , where t is the thickness of the pipe, 
and j?, the tensile stress per unit area induced by the end-pres- 
sures (fluid). 




536 



MECHANICS OF ENGINEEKING. 



For equilibrium, therefore, we may put ^(hor. comps.) = ; 
i.e., 

7tr 2 p — nrtya — 2nrtp x = ; 



JPi = 



_ r(j> -p a ) 



2t 



(1) 



(Strictly, the two circular areas sustaining the fluid pressures 
are different in area, but to consider them equal occasions but 
a small error.) 

Eq. (1) also gives the tension in the central section of a thin 
hollow sphere, under bursting pressure. 

Case II. Stresses in the longitudinal section of pipe, due to 
radial fluid pressures .* — Consider free the half (semi-circular) 

of any length I of the pipe, be- 
tween two cross-sections. Take an 
axis X (as in Fig. 470) "] to the 
longitudinal section which has been 
made. Let #„ denote the tensile 

a. •air..// /.:.'. .'],;:;•■ '! ! . 

j'/.'-y,'/ i. - '1 ' •'-•. v5*4-^x, s ^ r ess (per unit area) produced in 

the narrow rectangles exposed at A 
and B (those in the half-ring edges, 
having no X components, are not 
drawn in the figure). On the in- 
ternal curved surface the fluid pres- 
sure is considered of equal intensity 
= p at all points (practically true even with liquids, if 2r is 
small compared with the head of water producing p). The 
fluid pressure on any dF or elementary area of the internal 
curved surface is = pdF. Its X component (see § 416) is 
obtained by multiplying^? by the projection of dF on the ver- 
tical plane ABC, and since p is the same for all the dF's of 
the curved surface, the sum of all the ^components of the in- 
ternal fluid pressures must = p multiplied by the area of rect- 
angle ABCD, = 2rlp ; and similarly the X components of the 




Fig. 470. 



* Analytically this problem is identical with that of the smooth cord on 
a smooth cylinder, § 169, and is seen to give the same result. 



BUESTING OF PIPES. 537 

external atmos. pressures = 2rlp a (nearly). The tensile stresses 
( || to X) are equal to 2ltp 3 ; hence for equilibrium, 2X = 
gives 

2ltp, - 2rlp + 2rlp a = ; 

.:* = &=** (2) 

This tensile stress, called hoop tension, p^, opposing rupture by 
longitudinal tearing, is seen to be double the tensile stress p 1 
induced, under the same circumstances, on the annular cross' 
section in Case I. Hence eq. (2), and not eq. (1), should be 
used to determine a safe value for the thickness of metal, t, or 
any other one unknown quantity involved in the equation. 

For safety against rupture, we must put p 2 = T ' , a safe 
tensile stress per unit area for the material of the pipe or tube 
(see §§ 195 and 203) ; 

,. t = -l£^\. ..... (3) 

(For a thin hollow sphere, t may be computed from eq. (1) ; 
that is, need be only half as great as with the cylinder, other 
things being equal.) 

Example. — A pipe of twenty inches internal diameter is to 
contain water at rest under a head of 340 feet ; required the 
proper thickness, if of cast-iron. 

340 feet of water measures 10 atmospheres, so that the in 
ternal fluid pressure is 11 atmospheres ; but the external pres 
sure^? a being one atmos., we must write (inch, lb., sec.) 

(P—Pa) = 10 X 14.7 = 147.0 lbs. per sq. in., and r = 10 in., 

while (§ 203) we may put T' = i of 9000 = 4500 lbs. per sq 
in. ; whence 

10 v 147 
t = ** = 0.326 inches. 

4500 



538 MECHANICS OF ENGINEERING. 

But to insure safety in handling pipes and imperviousness to 
the water, a somewhat greater thickness is adopted in practice 
than given by the above theory. 

Thus, "Weisbach recommends (as proved experimentally also) 
for 

Pipes of sheet iron, t = [0.00172 rA + 0.12] inches; 
" " cast " t = [0.00476 rA + 0.34] « 
" " copper t = [0.00296 rA 4- 0.16] " 
" " lead t = [0.01014 rA + 0.21] « 

" " zinc t = [0.00484 rA + 0.16] " 



A 



in which t = thickness in inches, r = radius in inches, and A 
= excess of internal over external fluid pressure (i.e., p — p a ) 
expressed in atmospheres. 

For instance, for the example just given, we should have 
(cast-iron) 

t = .00476 X 10 X 10 + 0.34 = 0.816 inches. 

If the pipe is subject to " water-ram" (§ 422) the strength 
should be much greater. To provide against " water-ram," 
Mr. J. T. Fanning, on p. 453 of his " Hydraulic and Water- 
supply Engineering," advises adding 230 feet to the static 
head in computing the thickness of cast-iron pipes. 

For thick hollow cylinders see Rankine's Applied Mechan- 
ics, p. 290, and Cotterill's Applied Mechanics, p. 403. 

427. Collapsing of Tubes under Fluid Pressure. (Cylindrical 
boiler-flues, for example.) — If the external exceeds the internal 
fluid pressure, and the thickness of metal is small compared 
with the diameter, the slightest deformation of the tube or 
pipe gives the external pressure greater capability to produce 
a further change of form, and hence possibly a final collapse; 
just as with long columns (§ 303) a slight bending gives great 
advantage to the terminal forces. Hence the theory of § 426 
is inapplicable. According to Sir ¥m. Fairbairn's experi- 
ments (1858) a thin wrought-iron cylindrical (circular) tube 
will not collapse until the excess of external over internal 
pressure is 



COLLAPSE OF TUBES. 539 

t 

p(in lbs. per sq. in.) = 9672000 — . . .(1). . (not homog.) 

Let 

(t. I. and d must all be expressed in the same linear unit.) 
Here t = thickness of the wall of the tube, d its diameter, and 
I its length ; the ends being understood to be so supported aa 
to preclude a local collapse. 

Example. — With I = 10 ft. = 120 inches, d = 4 in., and t = 
fa inch, we have 

p = 9672000 RA- -T- (120 X 4)1 = 201.5 lbs. per sq. inch. 

For safety, ■$■ of this, viz. 40 lbs. per sq. inch, should not be 
exceeded ; e.g., with 14.7 lbs. internal and 54.7 lbs. external. 

[Note. — For simplicity the power of the thickness used in eq. (1) above 
has been given as 2.00. In the original formula it is 2.19, and then all 
dimensions must be expressed in inches. A discussion of the experiments 
of Mr. Fairbairn will be found in a paper read by Prof. Unwin before the 
Institute of Civ. Engineers (Proceedings, vol. xlvi.). See also Prof. Unwin 's 
" Machine Design," p. 66. It is contended by some that in the actual con- 
ditions of service, boiler-flues are subjected to such serious straining- 
actions due to unequal expansion of the connecting parts as to render the 
above formula quite unreliable, thus requiring a large allowance in its 
application.] 



MO MECHANICS OF ENGINEERING. 



CHAPTEE II. 

HYDROSTATICS (Continued)— PRESSURE OF LIQUIDS IN TANKi 
AND RESERVOIRS. 

428. Body of Liquid in Motion, but in Relative Equilibrium.— 
By relative equilibrium it is meant that the particles are not 
changing their relative positions, i.e., are not moving among 
each other. On account of this relative equilibrium the fol- 
lowing problems are placed in the present chapter, instead of 
under the head of Hydrodynamics, where they strictly belong. 
As relative equilibrium is an essential property of rigid bodies, 
we may apply the equations of motion of rigid bodies to bodies 
of liquid in relative equilibrium. 

Case I. All the particles moving in parallel right lines 
with equal velocities ; at any given instant (i.e., a motion of 
translation.) — If the common velocity is constant we have a 
uniform translation, and all the forces acting on any one par- 
ticle are balanced, as if it were not moving at all (according to 
Newton's Laws, § 54); hence the relations of internal pressure, 
free surface, etc., are the same as if the liquid were at rest. 
Thus, Fig. 471, if the liquid in the moving tank is at rest rel- 
v atively to the tank at a given instant, with 
its free surface horizontal, and the motion 
of the tank be one of translation with a uni- 
form velocity, the liquid will remain in this 
condition of relative rest, as the motion 
proceeds. 

But if the velocity of the tank is accelerated with a constant 
acceleration = p (this symbol must not be confused with p 
for pressure), the free surface will begin to oscillate, and finally 
come to relative equilibrium at some angle a with the horizon- 
tal, which is thus found, when the motion is horizontal. See 
Fig. 4T2, in which the position and value of a are the same, 
whether the motion is uniformly accelerated from left to right 




EELATIVE EQUILIBRIUM OF LIQUIDS. 



541 




or uniformly retarded from right to left. Let be the lowest 

point of the free surface, and Ob a v n > P 

small prism of the liquid with its 

axis horizontal, and of length = x ; 

nb is a vertical prism of length = *~T~ 

z y and extending from the extremity 

of Ob to the free surface. The TVT 

pressure at both and n is p a = #^«fc 

atmos. pres. Let the area of cross- te 472# 

section of both prisms be = dF. 

Now since Ob is being accelerated in direction ^(horizont.)^ 
the difference between the forces on its two ends, i.e., its ^X y 
must = its mass X accel. (§ 109). 

.-. p b dF - jp a dF = [xdF. y + g\p. . . . (1) 

(y = heaviness of liquid ; p h = press, at b) ; and since the ver- 
tical prism nb has no vertical acceleration, the ^(vert. com- 
pons.) for it must = (X 

.\p b dF-p a dF-zdF.y=0 (2) 

From (1) and (2), 

xy — z p 

-L.p = zy: ,\ — =*-. 

g x g 

Hence On is a right line, and therefore 



(3) 



tan or, or — , 

x 



9' 



(4) 



[Another, and perhaps more direct, method of deriving this 
result is to consider free a small particle of the liquid lying in 
the surface. The forces acting on this particle are two : the 
first its weight = dG ; and the second the resultant action of 
its immediate neighbor-particles. Now this latter force (point- 
ing obliquely upward) must be normal to the free surface of 
the liquid, and therefore must make the unknown angle a with 
the vertical. Since the particle has at this instant a rectilinear 
accelerated motion in a horizontal direction, the resultant of the 
two forces mentioned must be horizontal and have a value = 
mass X acceleration. That is, the diagonal formed on the two 



542 



MECHANICS OF ENGINEEKING. 



forces must be horizontal and have the value mentioned, = 
(dG -j- g)p ; while from the nature of the figure (let the stu- 
dent make the diagram for himself) it must also = dG tan a. 



Q. E. 



n J 



.\ dG tan a = — . p ; or, tan a=^-. . . 
9 9 

If the translation were vertical, and the acceleration upward 
[i.e., if the vessel had a uniformly accelerated upward motion 
or a uniformly retarded downward motion], the free surface 
would be horizontal, but the pressure at a depth = h below the 
surface instead of p =p a -\-? l y would be obtained as follows : 
Considering free a small vertical prism of height = h with 
upper base in the free surface, and putting ^(vert. compons.) 
= mass X acceleration, we have 

_ hdF. y 



dF.p-dF.p a -hdF.y 



9 



V 



if 



(5) 



If the acceleration is downward (not the velocity necessarily) 

we make^? negative in (5). If the vessel falls freely, p =— g 

and .'. p =p a , in all parts of the liquid. 

Query : Suppose p downward and > g. 

Case II. Uniform Rotation about a Vertical Axis. — If the 

narrow vessel in Fig. 473, open at top and containing a liquid, 

be kept rotating at a uniform angu- 
lar velocity go (see § 110) about a 
vertical axis Z, the liquid after some 
oscillations will be brought (by fric- 
tion) to relative equilibrium (rotat- 
ing about Z, as if rigid). Required 
the form of the free surface (evi- 
dently a surface of revolution) at 
each point of which we know 

Let be the intersection of the 
axis Z with the surface, and n any point in the surface ; b being 




Fig. 473. 



UNIFORM ROTATION- OF LIQUID IN VESSEL. 543 

a point vertically under n and in same horizontal plane as 0. 
Every point of the small right prism nb (of altitude = z and 
sectional area dF) is describing a horizontal circle about Z. and 
has therefore no vertical acceleration. Hence for this prism, 
free, we have 2Z — 0; i.e., 

dF.p h - dF.p a - zdF. r = (1) 

Now the horizontal right prism Ob (call the direction ...b, 
X) is rotating uniformly about a vertical axis through one ex- 
tremity, as if it were a rigid body. Hence the forces acting 
on it must be equivalent to a single horizontal force, — ca 9 J/p, 
(§122a,) coinciding in direction with X. [M= mass of prism 
=± its weight -5- g, and p = distance of its centre of gravity 
from ; here p = \x = £ length of prism]. Hence the ~2X 

of the forces acting on the prism Ob must = — go 2 y\x. 

But the forces acting on the two ends of this prism are their 
own ^components, while the lateral pressures and the weights 
of its particles have no Xcompons. ; 

.:dF.p a -dF.p b = L. . . (2) 

From (1) and (2) we have 

_(aw) , _«f. ,„x 

*~~W -^ () 

where v = gdx = linear velocity of the point n in its circular 
path. 

[As in Case I, we may obtain the same result by considering 
a single surface particle free, and would derive for the resultant 
force acting upon it the value dG tan a in a horizontal direc- 
tion and intersecting the axis of rotation. But here a is dif- 
ferent for particles at different distances from the axis, tan a 

dz 
being the — of the curve On. As the particle is moving uni- 

(XX 

formly in a circle the resultant force must point toward the 



544 MECHANICS OF ENGINEERING. 



9 



dC 
centre of the circle, i.e., horizontally, and have a value . - 

9 x 

where x is the radius of the circle [§ 74, eq. (5)] ; 

jn . dG (goxY dz oo % x 
•\ aijr tan a = - — J ; or tan #,=—-,= ; 

g x dx g 

,\ / dz = — / xdx: or, z = — . — . . Q. E. D. 

t/O gd0 > ' ^2 

Hence any vertical section of the free surface through the 
axis of rotation Z is a parabola, with its axis vertical and vertex 
at 0; i.e., the free surface is & paraboloid of revolution, with 
Z as its axis. Since gdx is the linear velocity v of the point 
b in its circular path, z = " height due to velocity" r [§ 52]. 

Example. — If the vessel in Fig. 473 makes 100 revol. per 
minute, required the ordinate z at a horizontal distance of a? = 
4 inches from the axis (ft.-lb.-sec. system). The angular veloc- 
ity go = \2n 100 •+- 60] radians per sec, [N. B. — A radian == 
the angular space of which 3.1415926 . . . make a half -re vol., 
or angle of 180°]. With x = i it and ^ = 32.2, 



and the pressure at b (Fig. 471) is (now use inch, lb., sec.) 
ifc =p a + zy= 147+ 2J- X yjg - 14.781 lbs. per sq. in. 

Prof. Mendelejeff of Russia has recently utilized the fact an- 
nounced as the result of this problem, for forming perfectly 
true paraboloidal surfaces of plaster of Paris, to receive by 
galvanic process a deposit of metal, and thus produce specula 
of exact figure for reflecting telescopes. The vessel contain- 
ing the liquid plaster is kept rotating about a vertical axis 
at the proper uniform speed, and the plaster assumes the de- 
sired shape before solidifying. A fusible alloy, melted, may ; 
also be placed in the vessel, instead of liquid plaster. • 



EELATIVE EQUILIBELUM. 



545 



,^ 



*>---. 



Eemahk. — If the vessel is quite full and closed on top, ex- 
except at 0' where it communicates 
by a stationary pipe with a reser- 
voir, Fig. 474, the free surface 
cannot be formed, but the pres- 
sure at any point in the water is 
just the same during uniform rota- 
tion, as if a free surface were formed 
with vertex at ; 

i.e., <p b = <p a + (A -f z)y. . (4) 

See figure for h and z. (In subse- 
quent paragraphs of this chapter 
the liquid will be at rest.) 




Fig. 474. 



428a. Pressure on the Bottom of a Vessel containing Liquid at 
Rest. — If the bottom of the vessel is plane and horizontal, the 
intensity of pressure upon it is the same at all points, being 



J -AIK?-."'V~--- , . 



^£k^ 



Fig. 475. 



•■'■■.-•AIR-v-');'- 



E 



$T 



^B 



Fig. 476. 



p=p a -{- hy (Figs. 475 and 476), and the pressures on the ele- 
ments of the surface form a set of parallel (vertical) forces. 
This is true even if the side of the vessel overhangs, Fig. 476, 
the resultant fluid pressure on the bottom in both cases being 



P = Fp-Fp a = Fhy. 



(1) 



(Atmospheric pressure is supposed to act under the bottom.) 
It is further evident that if the bottom is a rigid homogeneous 
plate and has no support at its edges, it may be supported at a 




546 MECHANICS OF ENGINEERING. 

single point (Fig. 477), which in this case (horizontal plate) 
is its centre of gravity. This point is called 
the Centre of Pressure, or the point of appli- 
cation of the resultant of all the fluid pressures 
acting on the plate. The present case is such 
that these pressures reduce to a single result- 
ant, but this is not always practicable. 

Example. — In Fig. 476 (cylindrical vessel 
fig. 4?7. containing Water), given h = 20 ft., h x = 

15 ft., r 1 = 2 ft., 7\ = 4 ft., required the pressure on the bot- 
tom, the vertical tension in the cylindrical wall CA, and the 
hoop tension (§ 426) at C. (Ft., lb., sec.) Press, on bottom = 
Fhy = nr^hy = TtlQ X 20 X 62.5 = 62857 lbs. ; while the 
upward pull on CA = 

(nr* _ nr?)h x y = tt(16 - 4)15 X 62.5 = 35357 lbs. 

If the vertical wall is t = ■£$ inch thick at C this tension will 
be borne by a ring-shaped cross-section of area = Inrj (nearly) 
= 2^-48 X T V = 30.17 sq. inches, giving (35357 -r- 30.17) = 
about 1200 lbs. per sq. inch tensile stress (vertical). 
The hoop tension at C is horizontal and is 

p" — rip —p a ) + t (see § 426), where p =p a + \y ; 

„ 48 X 15 X 12 X (62.5 + 1728) Q10K ., 
% \p" = --i i = 3125 lbs. per sq. in. 

TO 

(using the inch and pound). 

429. Centre of Pressure. — In subsequent work in this chapter, 
since the atmosphere has access both to the free surface of 
liquid and to the outside of the vessel walls, and^? a would can- 
cel out in finding the resultant fluid pressure on any elemen- 
tary area dF of those walls, we shall write : 

The resultant fluid pressure on any dF of the vessel wall is 
normal to its surf ace and is dP = pdF '= zydF, in which z 
is the vertical distance of the element below the free surface 
of the liquid (i.e., z = the " head of water"). If the surface 
pressed on is plane, these elementary pressures form a system 
of parallel forces, and may be replaced by a single resultant 



CENTRE OF PRESSURE. 



547 



(if the plate is rigid) which will equal their sum, and whose 
point of application, called the Centre of Pressure, may be 
located by the equations of § 22, put into calculus form. 

If the surface is curved the elementary pressures form a sys 
tem of forces in space, and hence (§ 38) cannot in general be 
reduced to a single resultant, but to two, the point of applica- 
tion of one of which is arbitrary (viz., the arbitrary origin, 
§ 38). 

Of course, the object of replacing a set of fluid pressures by 
a single resultant is for convenience in examining the equi- 
librium, or stability, of a rigid body the forces acting on which 
include these fluid pressures. As to their effect in distorting 
the rigid body, the fluid pressures must be considered in their 
true positions (see example in § 264), and cannot be replaced 
by a resultant. 

430. Resultant Liquid Pressure on a Plane Surface forming 
Part of a Vessel Wall. Co-ordinates of the Centre of Pressure. — 
Fig. 478. Let AB be a. portion (of any shape) of a plane 
surface at any angle with the 
horizontal, sustaining liquid 
pressure. Prolong the plane 
of AB till it intersects the free 
surface of the liquid. Take 
this intersection as an axis Y, 
being any point on Y. The 
axis X, ~| to Y, lies in the 
given plane. Let a = angle 
between the plane and the free 
surface. Then x and y are the 
co-ordinates of any elementary fig. 478. 

area dF of the surface, referred to Xand Y. z = the "head 
of water," below the free surface, of any dF. The pressures 
are parallel. 

The normal pressure on any dF = zydF; hence the sum of 
these, = their resultant, 




= F, = r fzdF=Fzy; 



(1) 



548 MECHANICS OF ENGINEERING. 

in which z = the u mean 2," i.e., the z of the centre of gravity 
G of the plane figure AB, and F — total area of AB \_F z == 
fzdF, from eq. (4), § 23]. y = heaviness of liquid (see § 409). 
That is, the total liquid pressure on a plane figure is equal 
to the weight of an imaginary prism of the liquid having a 
base = area of the given figure and an altitude == vertical 
depth of the centre of gravity of the figure below the surface 
of the liquid. For example, if the figure is a rectangle with 
one base (length = b) in the surface, and lying in a vertical 
plane, 

P~bh.\hy^ \bh'y. 

Evidently, if the altitude be increased, P varies as its square. 

From (1) it is evident that the total pressure does not de- 
pend on the horizontal extent of the water in the reservoir. 

Now let x c and y c denote the co-ordinates, in plane YOX, 
of the centre of pressure, C, or point of application of the re- 
sultant pressure P, and apply the principle that the sum of 
the moments of each of several parallel forces, about an axis "1 
to them, is equal to the moment of their resultant about the 
same axis [§22]. First taking OY as an axis of moments, 
and then OX, we have 

Px c = J^(zydF)x, and Py c = J B (zydF)y. . (2) 

But P = Fzy = Fx(sm a)y, and the z of any dF — x sin a. 
Hence eqs. (2) become (after cancelling the constant, y sin a) 

^JW^Ir^' &g£ . . (3) 

Fx Fx Fx 

in which I Y = the " mom. of inertia" of the plane figure re- 
ferred to Y (see § 85). [!N". B. — The centre of pressure as 
thus found is identical with the centre of oscillation (§ 117) 
and the centre of percussion [§ 113] of a thin homogeneous 
plate, referred to axes X and Y, Y being the axis of suspen- 
sion.] 

Evidently, if the plane figure is vertical a == 90°, x = z for 



CENTRE OF PRESSURE. 549 

all dF's, and x == z. It is also noteworthy that the position 
of the centre of pressure is independent of a. 

Note. — Since the pressures on the equal dF*B lying in any 
horizontal strip of the plane figure form a set of equal parallel 
forces equally spaced along the strip, and are therefore equiva- 
lent to their sum applied in the middle of the strip, it follows 
that for rectangles and triangles with horizontal bases, the 
centre of pressure must lie on the straight line on which the 
middles of all horizontal strips are situated. 

431. Centre of Pressure of Rectangles and Triangles with Bases 
Horizontal. — Since all the dF'& of one horizontal strip have 
the same x, we may take the area of the strip K 

for dF in the summation fx^dF. Hence for 
the rectangle AB, Fig 479, we have from eq. 
(3), §430, mthdF=zbdx, 



h 



^xydx _ 2 K _ K 



: A 

hi 

i 



■i — *■ 



i— '-- 
c. 



dx 



0{h Q —/l 1 ) — - — Fig. 479. 

while (see note, § 430) y c = \b. 

When the upper base lies in the surface, h x — 0, and x = 
|A 2 =fqfthe altitude. 

For a triangle with its base horizontal and vertex up, Fig. 
480, the length w of a horizontal strip is variable and dF— 

udx. From similar triangles u = — —{x — h) ; therefore 

th n th. 



*h 



I x'dF , =- / x*(x — h,)dx 



Fx y>{k~h)\.h+i{h-ky\ 

■Bnt£xXx-h)dx=^-hf) 
= 1 V(3A,' + V-4AA') 

3V + 2A A+V ,. 

■•*'-* 2A7+1, — (2) 



550 



MECHANICS OF ENGINEERING. 



Also, since the centre of pressure must lie on the line AB join~ 
ing the vertex to the middle of base (see note, § 430), we easily 
determine its position. 

Evidently for h 1 = 0, i.e., when the vertex is in the- surface, 




x c = f A 2 . Similarly, for a triangle with 
base horizontal and vertex dotvn, Fig. 481, 
we find that 

i sh; + 2hA + K 

2 " 



x = — 



(3) 



2/,, + K 

If the base is in the surface, A 1 = and 
(3) reduces to x c = \h r 

It is to be noticed that in the case of the triangle the value 
of x c is the same whatever be its shape, so long as A, and A 2 
remain unchanged and the base is horizontal. If the base is 
not horizontal, we may easily, by one horizontal line, divide 
the triangle into two triangles whose bases are horizontal and 
whose combined areas make up the area of the first. The re- 
sultant pressure on each of the component triangles is easily 
found by the foregoing principles, as also its point of applica- 
tion. The resultant of the two parallel forces so determined 
will act at some point on the line joining the centres of pres- 
sures of the component triangles, this point being easily found 
by the method of moments, while the amount of this final re- 
sultant pressure is the sum of its two components, since the 
latter are parallel. An instance of this procedure will be 
given in Example 3 of § 433. Similarly, the rectangle of Fig. 
479 may be distorted into an oblique parallelogram with hori- 
zontal bases without affecting the value of x c , nor the amount 
of resultant pressure, so long as h x and A 3 remain unchanged. 

432. Centre of Pressure of Circle. — Fig. 482. It will lie on 
the vertical diameter. Let r = radius. From eq. (3), § 413, 




r x 



_ I g +Fx __ JTtr'+Trr'x' 
F x nr 2 x 



(See eq. (4), § 88, and also § 91.) 
- , 1 r a 

.*. X = X + T * = ' 



W 



CENTEE OF PEESSUEE. 



551 



433. Examples. — It will be noticed that although the total 
pressure on the plane figure depends for its value upon the 
head, 1:, of the centre of gravity, its point of application is al- 
ways lower than the centre of gravity. 

Example 1. — If 6 ft. of a vertical sluice-gate, 4 ft. wide, 
Fig. 483, is below the water-surface, the total 
water pressure against it is (ft, lb., sec. ; eq. 
(1), § 430) 



IN. 



P = Fzy = 6 X 4 X 3 X 62.5 = 4500 lbs., 



* — *r 



4— *C 



Fig. 483. 



and (so far as the pressures on the vertical 
posts on which the gate slides are concerned) 
is equivalent to a single horizontal force of 
that value applied at a distance x a = § of 
6 = 4 f t. below the surface (§ 431). 

Example 2. — To (begin to) lift the gate in Fig. 483, the 
gate itself weighing 200 lbs., and the coefficient of friction 
between the gate and posts being/" = 0.40 (abstract numb.) (see 
§ 156), we must employ an upward vertical force at least 



= P' = 200 + 0.40 X 4500 = 2000 lbs. 



Example 3. — It is required to find the resultant hydro- 
static pressure on the trapezoid in Fig. 483& with the dimen- 
sions there given and its bases horizontal ; also its point of ap- 
plication, i.e., the centre of pressure of 
the plane figure in the position there 
D shown. From symmetry the C. of P. will 
be in the middle vertical of the figure, 
as also that of the rectangle B OFF, and 
that of the two triangles ABE and 
CDF taken together (conceived to be 
shifted horizontally so that OF and 
BE coincide on the middle vertical, 
thus forming a single triangle of 5 ft. base, and having the 
same total pressure and C. of P. as the two actual triangles 
taken together). Let P 1 — the total pressure, and xj refer to 
the C. of P., for the rectangle ; P a and a?/, for the 5 ft. tri- 











W 




A B 


i 


C D 


\| 


q' 


1/ 


\| 


i 


!/ 


E 




F 


AD--.IO' 






EF- 5' 






Fig. 


483a. 





552 



MECHANICS OF ENGINEEKING. 



angle; \ = 4 ft. and k, = 10 ft. being the same for both. 
Then from eq. (1), § 430, we have (with the ft., lb., and sec.) 

P 1 =S0x7y=2l0y; and P, = i X 6 X 5 X 6y = 90 r ; 

while from eqs. (1) and (3) of § 431 we have also (respectively) 



x/ = 



x/ = 



2 1000 - 64 2 936 



3 ' 100 ~ 16 



84 



= 7.438 feet; 



r 80 + 1 00 228 

8+10 " 2X18 



= 6.333 feet. 



The total pressure on the trapezoid, being the resultant of 
P x and P 2 , has an amount = P x -f P^ (since they are parallel), 
and has a lever-arm x c about the axis OY to be found by the 
principle of moments, as follows : 



x = P& + P&: = (210X 7.438 + 90 X 6.33) r 
P, + -P, (210 + 90> 



7.09 ft. 



The total hydrostatic pressure on the trapezoid is (for fresh 
water) 

P = P 1 +P 2 = [210 + 90] 62.5 = 18750 lbs. 

Example 4.— Eequired the horizontal force P\ Fig. 484, to 
be applied at N (with a leverage of a' = 30 inches about the 

fulcrum M) necessary to (begin 
to) lift the circular disk AB of 
radius r == 10 in., covering an 
opening of equal size. JSfMAB 
is a single rigid lever weighing 
G' =p 210 lbs. The centre of 
gravity, G, of disk, being a ver- 
tical distance z = O' G = 40 
inches from the surface, is 50 
inches (viz., the sum of OM = 
k = 20" and MG = 30") from 
axis OY; i.e., x = 50 inches. 
The centre of gravity of the 
whole lever is a horizontal distance b', — 12 inches, from Jf. 




Fig. 484. 



EXAMPLES — CENTKE OF PKESSUKE. 553 

For impending lifting we must have, for equilibrium of the 
lever, 

P'a / =G / b / + P(x c -k); . . . . (1) 

where P = total water pressure on circular disk, and x = 
00. From eq. (1), § 430, (using inch, lb., and sec.,) 

P = Fzy = nr'zy = ttIOO X 40 X -£— = 454.6 lbs. 

' 1728 

_ 1 r 2 1 100 

From §432, x c = 00 = x + ±i = 50 + ± . =^ = 50.5 in. 

4: x ^ OU 

= ^ t 210 X 12 + 454.6 x 30.5] = 546 lbs. 

434. Example of Flood-gate. — Fig. 485. Supposing the rigid 
double gate AJD, 8 ft. in total width, to 
have four hinges ; two at 0, and two aty j 



1 ft. from top and bottom of water chan- "^eeFIP^L^: , 

nel ; required the pressures upon them, ==+ \t i 2 

taking dimensions from the figure (ft., £E | ! L. 

lb., sec). — -]-= i > 



13 ti 



Wat. press. = i 3 = F zy 

= 72 X 4J X 62.5 = 20250 "ISTX 

pounds, and its point of application (cent, of press.) is a dis- 
tance x c = § of 9' = 6' from 6> (§ 431). Considering the 
whole gate free and taking moments about e, we shall have 

(press, at f)xT = 20250 X 5 ; .-. press, at / = 14464 lbs. 

{half on each hinge aty), and 

.*. press, at e = P — press, aty = 5875 lbs. 

(half coming on each hinge). 



MECHANICS OF ENGINEERING. 



If the two gates do not form a single rigid body, and hence 
are not in the same plane when closed, a wedge-like or toggle- 
joint action is induced, producing much greater thrusts against 
the hinges, and each of these thrusts is not "1 to the plane of 
the corresponding gate. Such a case forms a good exercise 
for the student. 

435. Stability of a Vertical Rectangular Wall against Water 
Pressure on One Side. — Fig. 486. All dimensions are shown in 



«~V- 










■g' 


A 


=T*=' 


i 




= i— 


Z— 


| 


Eiftjfr 


z c =n 


i< 


-1 — 


r -. 
=p — 






— i i 


B 


E 

\e'I 




— | — 


— — 




v — — 


y- 


;':■'■.'■-:■'■'■'■':■ 


SKf 


■. .-.' - ?• > 


L__^ 


D 



Fig. 486. 



the figure, except I, which is the length 
of wall "1 to paper. Supposing the wall 
to be a single rigid block, its weight G' 
= h'h'ly' (y' being its heaviness (§ 7), 
and I its length). Given the water 
depth = A, required the proper width 
V for stability. For proper security : 
First, the resultant of G' and the 
water-pressure P must fall within the 

base BD (or, which amounts to the same thing), the moment 

of G' about D, the outer toe of the wall, must be numerically 

greater than that of P ; and 

Secondly, P must be less than the sliding friction /6r' (see 

§ 156) on the base BD, 

Thirdly, the maximum pressure per unit of area on the 

base must not exceed a safe value (compare § 348). 

Now P — F~zy 



hi — y = -^hHy (y — heaviness of water) ; 

2 2i 



and x c = f h. 

Hence for stability against tipping about D, 

P\h must he < G'\V ; i.e., \tily < ^Wh'ly' ; . (1) 

while, as to sliding on the base, 

P must he <fG'\ i.e., \ttly <f h'h'ly'. . . (2) 

As for values of the coefficient of friction,/*, on the base of 
wall, Mr. Fanning quotes the following among others, from 
various authorities : 



STABILITY OF WALL. 



555 



For point- 


■dressed 


granite 


on dry clay, / = 


= 0.51 


ii ii 


a 


a 


" moist clay, 


0.33 


ii ii 


a 


a 


" gravel, 


0.58 


ii ii 


a 


a 


" smooth concrete, 


0.62 


ii ii 


a 


a 


" similar granite, 


0.70 


For dressed hard limestone on like limestone, 


0.38 


a 


t< 


a 


" brickwork, 


0.60 


For common bricks on common bricks, 


0.64 



To satisfactorily investigate the third condition requires the 
detail of the next paragraph. 



M ^-' ! \E P 




nb' 



Fig. 487. 



— > 



436. Parallelopipedical Keservoir Walls. More Detailed and 
Exact Solution. — If (1) in the last paragraph were an exact 
equality, instead of an inequality, 
the resultant R of P and G' 
would pass through the corner 
D, tipping would be impending, 
and the pressure per unit area at 
D would be theoretically infinite. 
To avoid this we wish the wall 
to be wide enough that the re- 
sultant R, Fig. 487, may cut 
BD in such a point, E\ as to cause the pressure per unit area, 
p m , at D to have a definite safe value (for the pressure p m at 
D, or quite near D, will evidently be greater than elsewhere 
on BD ; i.e., it is the maximum pressure to be found on BD). 
This may be done by the principles of §§ 346 and 362. 

First, assume that R cuts BD outside of the middle third; 
i.e., that 



'— *r 



VE\ = rib', > $'(pr n > 



i); 



where n denotes the ratio of the distance of E' from the mid- 
dle of the base to the whole width, h\ of base. Then the pres- 
sure (per unit area) on small equal elements of the base BD 
(see § 346) may be considered to vary as the ordinates of a 
triangle MND (the vertex M being within the distance BD), 
md E^D will = ±MD ; i.e., 



£56 MECHANICS OF ENGINEERING. 

The mean pressure per unit area, on MP, 



= G'-t-(I.MD), 

and hence the maximum pressure (viz., at D), being double 
the mean, is 

p m = %G> +{ZVl(i -»)]; . . . . (0) 

and if p m is to equal <7'(see §§ 201 and 203), a safe value for the 
crushing resistance, per unit area, of the material, we shall 
have 

b'l(& ~n)C / = %G , = \Vh'ly', 

1 2 Ay. m 

To find V, knowing n, we put the 2(moms.) of the G r and P 
at E, about E 7 ,= zero (for the only other forces acting on 
the wall are the pressures of the foundation against it, along 
MD ; and since the resultant of these latter passes through E\ 
the sum of their moments about E ! is already zero) ; i.e., 

G'nV - P\Ji = ; or, nb'*h'ly'= ±h . \~ttly ; 



V Znh'y' 



* = A-A s^r <») 



Having obtained b\ we must also ascertain if P is </G\ the 
friction ; i.e., if P is < fb'h'ly'. If not, V must be still further 
increased. (Or, graphically, the resultant of G' and P must 
not make an angle > 0, the angle of friction, with the ver- 
tical. 

If n, computed from (1), should prove to be < -J-, our first 
assumption is wrong, and we therefore assume n < -J-, and pro- 
ceed thus : 

Secondly, n being < \ (see §§ 346 and 362), we have a 



STABILITY OF RESERVOIE WALLS. 557 

trapezoid of pressures, instead of a triangle, on BD. Let the 
pressure per unit area at D be jp m (the maximum on base). 
The whole base now receives pressure, the mean pressure (per 
unit area) being = G' -f- \b'l\ ; and therefore, from § 362, 
Case I, we have 

fl. = t*» + l]|Ji (° a > 

and since, here, G' = Vh'ly\ we may write 
Fm = {Gn + l)hy. 

For safety as to crushing resistance we put 

6(n + l)h'y' = C'\ whence n = ^ ^7-7 - 1J . . (la) 

Having found ^ from eq. (la), we determine the proper 
width of base V from eq. (2), in case the assumption n < \ is 

verified. 

Example.— In Fig. 486, let N = 12 ft, h = 10 ft., while 
the masonry weighs (/ =) 150 lbs. per cub. ft. Supposing 
it desirable to bring no greater compressive stress than 100 lbs. 
per sq. inch (== 14400 lbs. per sq. ft.) on the cement of the 
joints, we put C = 14400, using the ft.-lb.-sec. system of units. 

Assuming n > -§-, we use eq. (1), and obtain 

_ 1 _ 2 12 X 150 _ 5^ 
n ~2 3' 14400 _ 12' 

which is > -J- ; hence the assumption is confirmed, also the 
propriety of using eq. (1) rather than (la). 
Passing to eq. (2), we have 

/ 62.5 X 10 ' „ , 

A' — 10 y a / — — 3.7 feet. 

b _ iu x y f x 12 x lg0 

But, as regards frictional stability, we find that, with/= 0.30, 
a low value, and V — 3.7 ft. (ft., lb., sec), 



558 MECHANICS OF ENGINEERING. 

P \m y 100 X 62.5 



fG' fb'h'ly' 2 X 0.3 X 3.7 X 12 X 150 



1.5; 



which is greater than unity, showing the friction to be insuf- 
ficient to prevent sliding (with y=0.30); a greater width 
must therefore be chosen, for frictional stability. 

If we make n = ^-, i.e., make P cut the base at the outer 
edge of middle third (§ 362), we have, from eq. (2), 



V = 10 X 



JJ^^ = 5.89 feet; 
V £ X 12 X 150 



and the pressure at D is now of course well within the safe 
limit ; while as regards friction we find 

P ^fG' = 0.92, < unity, 

and therefore the wall is safe in this respect also. 

With a width of base = 3.7 feet first obtained, the portion 
3ID, Fig. 487, of the base which receives pressure [according 
to Navier's theory (§ 346)] would be only 0.92 feet in length, 
or about one sixth of the base, the portion BM tending to 
open, and perhaps actually suffering tension, if capable (i.e., if 
cemented to a rock foundation), in which case these tensions 
should properly be taken into account, as with beams (§ 295), 
thus modifying the results. 

It has been considered safe by some designers of high 
masonry dams, to neglect these possible tensile resistances, as 
has just been done in deriving V = 3.7 feet ; but others, in 
view of the more or less uncertain and speculative character of 
Navier's theory, when applied to the very wide bases of such 
structures, prefer, in using the theory (as the best available), 
to keep the resultant pressure within the middle third at the 
base (and also at all horizontal beds above the base), and thus 
avoid the chances of tensile stresses. 

This latter plan is supported by Messrs. Church and Fteley, 
as engineers of the proposed Quaker Bridge Dam in connec- 
tion with the New Croton Aqueduct of New York City, in 
their report of 1887. See § 438. 



RESERVOIR WALLS. 



559 



437. Wall of Trapezoidal Profile. Water-face Vertical. — 

Economy of material is favored by using a trapezoidal profile, 
Fig. 488. With this form the 
stability may be investigated in 
a corresponding manner. The 
portion of wall above each 
horizontal bed should be ex- 
amined similarly. The weight 
G' acts through the centre of 
gravity of the whole mass. 

Detail.— -Let Fig. 488 show 
the vertical cross-section of a 

trapezoidal wall, with notation '\ •;•;•';•.;'«_„__&: __!_"_:_■_'.* 

for dimensions as indicated ; the FlG - 488 - 

portion considered having a length = I, ~] to the paper. Let 
y = heaviness of water, y' that of the masonry (assumed homo- 
geneous), with n as in § 436. 

For a triangle of pressure, MD, on the base, i.e., with n > -J-, 
or resultant falling outside the middle third (neglecting pos- 
sibility of tensile stresses on left of M\ if the intensity of 
pressure ^ m at D is to = C (§ 203), we put, as in § 436, 




b'lft -n}0' = W, i-e., = W • W + V'V> 



whence 



_l_lAy V + V 
n ~2 3C ' V ' 



(1)' 



For a trapezoid of pressure, i.e. with n < -J-, or the resultant 
of P and G' falling within the middle third, we have, as be- 
fore (§ 362, Case I), 



G' 



whence 



VTC'Vl 

n = ±-G 






. (la)' 



560 MECHANICS OF ENGINEERING. 

From the geometry of the figure, having joined the middles 
of the two bases, we have 

(§ 26,Prob. 6), and, by similar triangles, OV : XV :: gO : h' y 
whence 

y i w 



■-. gy, = oT+ m = I . <y + fW r »"> + «r. . (ay 

The lines of action of G' and P meet at E, and their results 
ant cuts the base in some point E' . The sum of their moments 
about E' should be zero, i.e., P .±k= G' . OE'\ that is, (see 
eq. (a) above, and eq. (1), § 430,) 

i.e., cancelling, 

h 3 r =ih'y'[(b' + 2&")(&' - I") -f 6?ib'(b' + b")\ (®f 

Hence we have two equations for finding two unknowns 
viz.: (l)'and (2)' when n > ■$- ; and (la)' and (2)' when n < \. 

For dams of small height (less than 40 ft., say), if we im- 
mediately put n — \, thus restricting the resultant pressure to 
the edge of middle third, and solve (2)' for b\ b" being as- 
sumed of some proper value for a coping, foot-walk, or road- 
way, while N may be taken enough greater than h to provide 
against the greatest height of waves, from 2.5 to 6 ft., the 
value of jp m at D will probably be < C . In any case, for a 
value of n =, or <, \ we put p m for Cm equation (la/ and 
solve for jp m , to determine if it is no greater than C. 

Mr. Fanning recommends the following values for C (in lbs. 



KESEKVOIK WALLS. 561 

per sq.foot) with coursed rubble masonry laid in strong mor- 
tar: 



For Limestone. 


Sandstone. 


Granite. 


Brick. 


a = 50,000 


50,000 


60,000 


35,000 


Av. heaviness of 1 
the masonry in V 152 
lbs. per cub. ft. \ 


132 


154 


120 



As to frictional resistance, P must be <fG'\ i.e., 

Wr<fVrW + t>") (3)' 

If the base is cemented to a rock foundation with good 
material and workmanship throughout, Messrs. Church and 
Fteley (see § 436) consider that the wall may be treated as 
amply safe against sliding on the base (or any horizontal bed), 
provided the other two conditions of safety are already satis- 
fied. 

438. Triangular Wall with Vertical Water-face. — Making 
h" = in the preceding article, the trapezoid becomes a right 
triangle, and the equations reduce to the following : 

*» = 3=%; for * > *> • • • • W 

and 

p m = $h'y' [6n + 1] for n < -§- . . . (la)" 

{p m not to exceed C in any case) ; while to determine the 
breadth of base, b', after n is computed [or assumed, for small 
height of wall], we have from eq. (2)', 

&y = £A'&V[6»+i] (2)" 

Also, for frictional stability, 

Wly must be < \fh'Vly' (3)" 




562 MECHANICS OF ENGINEEKING. 

439. High Masonry Dams. — Although the principle of the 
arch may be utilized for vertical stone dikes of small height 

(30 to 50 feet) and small span, for 
greater heights and spans the 
formula for hoop tension, § 426 (or 
rather, here, " hoop compression"), 
on the vertical radial joints of the 
horizontal arch rings, Fig. 489, calls 
for so great a radial thickness of 
joint in the lower courses, that 
straight dikes (or "gravity dams") 
are usually built instead, even 
where firm rock abutments are available laterally. 

For example, at a depth of 100 feet, where the hydrostatic 
pressure is hy = 100 X 62.5 = 6250 lbs. per sq. ft., if we as- 
sume for the voussoirs a (radial, horizontal) thickness = 4 ft., 
with a (horizontal) radius of curvature r = 100 feet, we shall 
find a compression between their vertical radial faces of (ft.> 
lb., sec.) 

„ rip — Pa) 100 X 6250 i KCOKA lu «. 

rp" = -±L — £21 — :_ — 156250 lbs. per sq. ft., 

or 1085 lbs. per sq. inch ; far too great for safety, even if there 
were no danger of collapse, the dike being short. If now the 
thickness is increased, in order to distribute the pressure over 
a greater surface, we are met by the fact that the formula for 
" hoop compression" is no longer strictly applicable, the law of 
distribution of pressure becoming very uncertain; and even 
supposing a uniform distribution over the joint, the thickness 
demanded for proper safety against crushing is greater than 
for a straight dam ^gravity dam") at a very moderate depth 
below the water surface, unless the radius of curvature of arch 
can be made small. But the smaller the radius the more does 
the dam encroach on the storage capacity of the reservoir, while 
in no case, of course, can it be made smaller than half the span. 
Another point is, that as masonry is not destitute of elas- 
ticity, the longer the span the more unlikely is it that the 
parts of the arch will " close up" properly, and develop the 



HIGH MASONRY DAMS. 563 

abutment reactions when the water is first admitted to the 
reservoir ; which should occur if it is to act as an arch instead 
of by gravity resistance. 

For these reasons the engineers of the proposed Quaker 
Bridge Dam reported unfavorably to the plan of a curved de- 
sign for that structure, and recommended that a straight dam 
be built. See reference in § 436. According to their designs 
this dam is to be 258 feet in height (which exceeds by about 90 
feet the height of any dam previously built), about 1400 feet 
in length at the top, and 216 feet in width at the lowest 
point of base, joining the bed-rock. 

More recently, however (1888), a board of experts, specially 
appointed for the purpose, having examined a number of dif- 
ferent plans, have reported favorably to the adoption of a 
.curved form for the dam, as offering greater resistance under 
extraordinary circumstances (impact of ice-floes, earthquakes, 
etc.), on account of its arched form (though resisting by 
gravity action under usual conditions) than a straight struc- 
ture ; and also as more pleasing in appearance. 

Fig. 490 shows the profile of a straight high masonry dam 
as designed at the present day. Assuming a width V = from 
6 to 22 feet at the top, and a sufficient h" (see figure) to ex- 
ceed the maximum height of Waves, the up-stream outline 
ACM is made nearly vertical and perhaps somewhat concave, 
while the down-stream profile BDN, by computation or 
graphical trial, or both, is so formed that when the reservoir is 
full the resultant i?, of the weight ^rr+ 

(t of the portion ABCD of ma- §h rr *J 
sonry above each horizontal bed, as EE^ 
CD, and the hydrostatic pressure P 
on the corresponding up-stream face 
AC, shall cut the bed CD in such a 
point E' as not to cause too great 
compression p m at the outer edge D 
(not over 85 lbs. per sq. inch accord- 
ing to M. Krantz in " Eeservoir FlG - 49 °- 
Walls"). p m being computed by one of the equations [(0) and 
(0a) of 1 436] 




564 MECHANICS OF ENGINEEEINGr. 

For E' outside the middle third ) _ 2G m 

and neglecting tension J ' '^ m ~ 3^ JJj) t ux_ n \* ' 

For E' inside middle third i . .p m = (^±il? ; n a y» 

* Jm UD.l V ; 

where Z = length of wall "1 to paper, usually taken = one foot, 
or one inch, according to the unit of length adopted ; for n, 
see §436. 

Nor, when the reservoir is empty and the water pressure 
lacking, must the weight G resting on each bed, as CD, cut 
the bed in a point E" so near the edge C as to produce exces- 
sive pressure there (computed as above). The figure shows 
the general form of profile resultiug from these conditions. 
The masonry should be of such a character, by irregular bond- 
ing in every direction, as to make the wall if possible a mono- 
lith. For more detail see next paragraph. 

440. Quaker Bridge Dam (on the New Croton Aqueduct). — 
Attempts, by strict analysis, to determine the equation of the 
curve DJV, AM being assumed straight, so as to bring the 
point E' at the outer edge of the middle third of its joint, or 
to make the pressure at D constant below a definite joint, have 
failed, up to the present time; but approximate and tentative 
methods are in use which serve all practical purposes. As an 
illustration the method set forth in the report on the Quaker 
Bridge Dam will be briefly outlined ; this method confines E' 
to the middle third. 

The width AB ~ h" is taken = 22 ; for a roadway, and h" — 
7 ft. The profile is made a vertical rectangle from A down 
to a depth of 33 ft. below the water surface {reservoir full). 
Combining the weight of this rectangle of masonry with the 
corresponding water pressure (for a length of wall == one foot), 
we find the resultant pressure comes a little within the outer 
edge of the middle third of the base of the rectangle, while 
p m is of course small. 

The rectangular form of profile might be continued below 
this horizontal joint, as far as complying with the middle 



QUAKER BRIDGE DAM. 565 

third requirement, and the limitation of pressure-intensity, is 
concerned ; but, not to make the widening of the joints too 
abrupt in a lower position where it would be absolutely re- 
quired, a beginning is made at the joint just mentioned by 
forming a trapezoid between it and a joint 11 ft. farther down, 
making the lower base of the latter of some trial width, which 
can be altered when the results to which it gives rise become 
evident. Having computed the weight of this trapezoid and 
constructed its line of action through the centre of gravity of 
the trapezoid, the value of the resultant G of this weight and 
that of the rectangle is found (by principle of moments or by 
an equilibrium polygon) in amount and position, and combined 
with the water pressure of the corresponding 44 ft. of water to 
form the force i?, whose point of intersection with the new 
joint or bed (lower base of trapezoid) is noted and the value of 
jp m computed. These should both be somewhat nearer their 
limits than in the preceding joint. If not, a different width 
should be chosen, and changed again, if necessary, until satis- 
factory. Similarly, another layer, 11 ft. in height and of 
trapezoidal form, is added below and treated in the same way ; 
and so on until in the joint at a depth of 66 ft. from the 
water surface a width is found where the point E' is very 
close upon its limiting position, while p m is quite a little under 
the limit set for the upper joints of the dam, 8 tons per square 
foot. For the next three 11 ft. trapezoidal layers the chief 
governing element is the middle-third requirement, E' being 
kept quite close to the limit, while the increase of p m to 7.95 
tons per sq. ft. is unobjectionable ; also, we begin to move 
the left-hand edge to the left of the vertical, so that when the 
reservoir is empty the point E" shall not be too near the up- 
stream edge C. 

Down to a depth of about 200 ft. the value of p m is allowed 
to increase to 10.48 tons per sq. ft., while the position of E f 
gradually retreats from the edge of its limit. Beyond 200 ft. 
depth, to prevent a rapid increase of width and consequent 
extreme flattening of the down-stream curve, p m is allowed 
to mount rapidly to 16.63 tons per sq. ft. (=231 lbs. per 
sq. in.), which value it reaches at the point iTof the base of 



566 MECHANICS OF ENGINEERING. 

the dam, which has a width = 216 ft., and is 258 feet below 
the water surface when the reservoir is full. 

The heaviness of the masonry is taken as y' = 156.25 lbs. 
per cubic foot, just f of y = 62.5 lbs. per cub. foot, the heavi- 
ness taken for water. 

When the reservoir is empty, we have the weight G of the 
superincumbent mass resting on any bed CD, and applied 
through the point E" ; the pressure per unit area at C can 
then be computed by eq. (la)'", § 439, n being the quotient of 
(iCD — CE") -=- Cl) for this purpose. In the present case 
we find E" to be within middle third at all joints, and the 
pressures at C to be under the limit. 

For further details the reader is referred to the report itself 
(reprinted in Engineering News, January, 1888, p. 20). The 
graphic results were checked by computation, Wegmann's 
method, applied to each trapezoid in turn. 

441. Earthwork Dam, of Trapezoidal Section. — Fig. 491. It is 
required to find the conditions of sta- 
bility of the straight earthwork dam 
ABDE, whose length =/, L to 
paper, as regards sliding horizontally 
on the plane AE; i.e., its frictional 
K — c--^» stability. With the dimensions of 
the figure, y and y' being the heavi- 
Fig. 491. nesses of the water and earth respec- 

tively (see § T), we have 

Weight of dam = G, = vol. X y' = lh\b + i(a, + c)]y\ (1) 

Eesultant water press. = P — Fzy == CA X I X %hy. . (2) 

Horiz. comp. of P = H = P sin a 

= [OA sin a~\\My = \ttly. ... (3) 

From (3) it is evident that the horizontal component of P is 
just the same, viz., = Til . \hy, as the water pressure would be 
on a vertical rectangle equal to the vertical projection of OA 



«— & — -> D 



,E 



EARTHWOKK DAM. 567 

and with its centre of gravity at the same depth (JA). Com- 
pare §416. Also, 

Vert. comp. of P = V = P cos a 

= [OA cos a\\Kly — iahly, ... (4) 

and is the same as the water pressure on the horizontal projec- 
tion of OA if placed at a depth = O'G = JA. 

For stability against sliding, the horizontal component of P 
must be less than the friction due to the total vertical pressure 
on the plane AE, viz., G 1 + V\ hence \if is the coefficient of 
friction on AJ?, we must have H <f \_G 1 -\- V], i.e. (see above), 



ffily must be < f Ih^b + i(a t + c)~\y' + icihly 



(5) 



However, if the water leak under the dam on the surface AE, 
so as to exert an upward hydrostatic pressure 

V = [a x + b + c]lhy, 
(to make an extreme supposition,) the friction will be only 

and (5) will be replaced by 

#</[#,+ F-n (6) 

Experiment shows (Weisbach) that with y=0.33 computa- 
tions made from (6) (treated as a bare equality) give satisfactory 
results. 

Example.— (Ft., 1-b., sec.) With / = 0.33, h = 20 ft., h x == 
22 ft., a = 24 ft., a, = 26.4 ft., and o = 30 ft., we have, mak- 
ing (6) an equality, with y' = 2y, 

£A7 K =/[y7A 1 (ft + ^±^ + i^AZ r - (a, + I + *)%] ; 

.-. i(400) = |[22(Z>+28.2)2+i(24x 20) -(26.4 + 6+30)20]; 
whence, solving for b, the width of top, b = 10.3 feet. 



568 



MECHANICS OF ENGINEERING. 



442. Liquid Pressure on Both Sides of a Gate or Rigid Plate. — 
The sluice-gate AB, for example, Fig. 492, receives a pressure, 

P x , from the " head-water" J/, and 

an opposing pressure P 2 from the 

n^ "tail-water" JV. Since these two 

■wss ^zTzz ^ horizontal forces are not in the same 

"°~-f=^ line, though parallel, their resultant 

:=zr R, which = P, — P 2 , acts horizon- 




~ ~ ^~ tally in the same plane, but at a dis- 

fig. 492. ' tance below 1 = u, which we may 

find by placing the moment of R about 1 , equal to the alge- 
braic sum of those of P, and P n about O t . 



.-. Ru = P x x ' - Plx c " + h). 
[P A VPK'+*)] 



u = 



P — P 



(1) 
(2) 



C 1 and <7 2 are the respective centres of pressure of the surfaces 
O x B and O^B, and u = distance of R from O x , while h = dif- 
ference of level between head and tail waters. If the surfaces 
O x B and 2 B are both rectangular, 



xj = %L and a?„ 



ih- 



Example. — Let the dimensions be as in Fig. 493, both sur- 
a faces under pressure being rect- 

% __ angular and 8 ft. wide. Then (ft., 

lb., sec.) R = P,-P,, or (§ 430) 

.[12X8X6-8X8X 4]62.5 

= 20000 lbs. = 10 tons; 

while from ex. (2) 

_ [12X8X6X8- 8 X8X 4(9|)]62.5 
20000 

That is, u = 6.93 feet, which locates C. Hence the pressure 
of the gate upon its hinges or other support is the same (aside 




WATEK ON BOTH SIDES OF GATE. 



from its own weight), provided it is rigid, as if the single 
horizontal force M = 10 tons acted at the point C, 2.93 ft. be- 
low the level of the tail-water surface. 

443. If the plate, or gate, is entirely below the tail-water 
surface, the resultant pressure is applied in the centre of gravity 
of the plate. — Proof as follows: Conceive the surface to be 
divided into a great number of small equal areas, each = dF; 
then, the head of water of any dF being = x x on the head- 
water side, and = a? 2 on the tail-water side, the resultant pres- 
sure on the dF is ydF{x^ — a? a ) — ykdF, in which h is the 
difference of level between head and tail water. That is, the 
resultant pressures on the equal dE's are equal, and hence 
form a system of equal parallel forces distributed over the plate 
in the same manner as the weights of the corresponding por- 
tions of the plate ; therefore their single resultant acts through 
the centre of gravity of the plate ; Q. E. D. This single re- 
sultant —fyhdF — yhfdF = Fhy. 

Example. — Fig. 494. The resultant pressure on a circular 
disk ab of radius = 8 inches, (in 
the vertical partition OK,) which 
has its centre of gravity 3 ft. ~ 

below the tail-water surface, with 
h = 2 ft., is (ft., lb., sec.) 



R = Fhy = Ttr'hy 

62.5 



— *■ 



= n 8 2 X 24 X 



1728 



= 174.6 lbs., 



b ' 



[A— a' 



m 

Fig. 494 



and is applied through the centre Jjw/ 
of gravity of the circle. Evi- 
dently B is the same for any 
depth below the tail-water surface, so long as h = 2 ft. 
the student find a graphic proof of this statement.] 



[Let 



444. Liquid Pressure on Curved Surfaces. — If the rigid surface 
is curved, the pressures on the individual dE's, or elements of 
area, do not form a system of parallel forces, and the single re- 
sultant (if one is obtainable) is not equal to their sum. In 



570 



MECHANICS OF ENGINEEKING. 



general, the system is not equivalent to a single force, but can 
always be reduced to two forces (§ 38) the point of application 
of one of which is arbitrary (the arbitrary origin of § 38) and 
its amount = V(2Xf + (2 Y) 2 + {2Z)\ 

A single Example will be given ; that of a thin rigid shell 
having the shape of the curved surface of a right cone, Fig. 
495, its altitude being h and radius of base = r. It has no 
bottom, is placed on a smooth horizontal table, vertex up, and 
is filled with water through a small hole in the apex 0, which is 

left open (to admit atmospheric 
pressure). What load, besides its 
own weight G\ must be placed 
upon it to prevent the water from 
lifting it and escaping under the 
edge A % The pressure on each 
dF of the inner curved surface is 
zydFsmd is normal to the surface. 
Its vertical compon. is zydF sin a, 
compon. == zydF cos a. The dF's have all 
j's (or heads of water). The lifting 




Fig. 495. 



and horizontal 
the same <*, but different 
tendency of the water on the thin shell is due to the vertical 
components forming a system of || forces, while the horizon- 
tal components, radiating symmetrically from the axis of the 
cone, neutralize each other. Hence the resultant lifting force 



is 



V= ^"(vert. comps.) = y sin afzdF = y sin a Fz ; (1) 



where F— total area of curved surface, and z — the "head of 
water" of its centre of gravity. Eq. (1) may also be written 
thus: 

V=yF~z-, . (2) 

in which F b = Fsin a = area of the circular base = area of 
the projection of the curved surface upon a plane ~| to the 
vertical, i.e., upon a horizontal plane. Hence we may write 



V=$ynr*h, 



(3) 



since z — fA, being the z of the centre of gravity of the curved 



CONICAL SHELL. 571 

surface and not that of the base, y = heaviness of water. If 
G' — weight of the shell and is < 7, an additional load of 

V — G' will be needed to prevent the lifting. If the shell has 
a bottom of weight = G", forming a base for the cone and 
rigidly attached to it, we find that the vertical forces acting on 
the whole rigid body, base and all, are: V upward; G' and 
G" downward; and the liquid pressure on the base, viz., 

V = nr^hy (§ 428^) also downward. Hence the resultant 
vertical force to be counteracted by the table is downward, and 

= G' + G" +V-V, which = G' + G" + ^nr'hy ; (4) 

i.e., the total weight of the rigid vessel and the water in it, as 
we know, of course, in advance. 



OHAPTEE III. 

EARTH PRESSURE AND RETAINING WALLS. 

[.Note. — This chapter was outlined and written mainly by 
Prof. C. L. Crandall, and is here incorporated with his permis- 
sion. The theory of earth pressure is arranged from Bau- 
meister.] 

445. Angle of Repose. — Granular materials, like dry sand, 
loose earth, soil, gravel, pease, shot, etc., on account of the 
friction between the component grains, occupy an intermediate 
position between liquids and large rigid bodies. When heaped 
up, the side of the mass cannot be made to stand at an inclina- 
tion with the horizontal greater than a definite angle called the 
angle of natural slope, or angle of repose, different for each 
material ; so that if the side of the mass is to be retained per- 
manently at some greater angle, a Retaining Wall (or "Revet- 
ment Wall" in military parlance) becomes necessary to sup- 
port it. If the material is somewhat moist it may be made to 
stand alone at an inclination greater than that of the natural 
slope, on account of the cohesion thus produced, but only as 
long as the degree of moisture remains ; while if much water 
is present, it assumes the consistency of mud and may require 
a much thicker wall, if it is to be supported laterally, than if 
dry. 

In dealing with earth to be supported by a retaining wall, 
we consider the former to have lost any original cohesion 
which may have existed among its particles, or that it will 
eventually lose it through the action of the weather ; and hence 
treat it as a granular material. 

A few approximate values of the angle of natural slope are 

572 






RETAINING WALLS. 



573 



given below, being taken from Fanning, p. 345 ; see reference 
on p. 538 of this work. 



Material. 



Dry sand, fine. . 
11 " coarse 

Damp clay 

Wet clay 

Clayey gravei. .. 

Shingle 

Gravel , 

Firm loam 

Vegetable soil. . . 
Peat 



Angle 
of Repose. 



28° 
30° 
45° 
15° 
45° 
42° 
38° 
36° 
35° 
20° 



Coefficient 
of Friction. 



.532 

.577 
L000 
.268 
L.OOO 
.900 
.781 
.727 
.700 
.364 



Ratio 
of Slope. 



Horiz. 

1.88 
1.73 
100 
3.73 
1.00 
1.11 
1.28 
1,38 
1.43 
2.75 



to ve 
to 



The angle of repose, or natural slope, is also, evidently, the 
angle of friction between two masses of the same granular 
material. 

446. Earth Pressure, and Wedge of Maximum Thrust. — Fig. 
496. Let AB be a retaining wall, having a plane face AB in 
contact with a mass of earth ABB, both wall and earth being 
of indefinite extent ~| to the paper. 

J^etAB be the natural slope of the earth, making an angle ft 
with the vertical (/? is the complement of the angle of repose ; 
see preceding table). Since AB, making an angle a with 
the vertical, is more nearly vertical than AB, the retaining 
wall is necessary, to keep the mass ABB in the position 
shown. The profile BOB may be of any form in this general 
discussion. Suppose the wall to be on the point of giving 
way ; then the following motions are impending : 

1st. Sliding is impending between some portion ABC A of 
the mass of earth and the remainder CAB, the surface of 
rupture AC {C not shown in figure because not found yet, 
but lying somewhere on the profile BCIB) being assumed 
plane, and making some angle <$' (to be determined) with the 
vertical. At this instant the resultant pressure N' of AG ' B 
on the plane AC of the mass ABC (a wedge) must make 
an angle = /3 ( = comp. of angle of friction) with AC on 
the upper side. 



574 



MECHANICS OF ENGINEEEING. 



2d. A downward sliding of the mass ABC along the back 
face AB of the wall. That is, the resultant pressure P' of 
the wall against the mass BA C at this instant makes an angle 




Fig. 496. 

8 (= complement of angle of friction between the earth and 
wall) with the plane AB and on the upper side. The weight 
of the wedge of earth BAG' will be called G', and we desire 
to find the pressure P' against the wall. 

Let BA (7 be a wedge (of the earth-mass), in which A C makes 
any angle with A V, and suppose it to be on the point of 
moving down and forcing out the wall; thus encountering 
friction both on the plane AC and the plane AB. Then the 
forces acting on it are three, acting in known directions ; viz. : 
6r, its own weight, vertical ; JV, the resultant pressure of the 
earth below it, making an angle (3 with AC on upper side; 
and P, the resultant pressure of the wall, at angle 6 with AB 
(see Fig. 496 for positions of JV and P). If now we express 
the force P in terms of and other quantities, and find that 
value 0', of 0, for which P is a maximum, we thereby deter- 
mine the "wedge of maximum thrust" ABC A ; while this 
maximum thrust, P\ is the force which the wall must be de- 
signed to withstand. [If the wall is overturned, the earth 
will sink with it until this part of its surface gradually as- 
sumes the natural slope.] 

Let G = weight of prism of base ABC, and altitude = unity 
"1 to paper; then G = y X area ABC, where y = " heavi- 
ness" = wgt. per cub. unit, of earth. Now P, G, and iV 
balance; therefore, in triangle abc, if ab and ac are drawn || 



EETAINING WALLS. 575 

and = G and irrespectively, bo is = and || to P\ and from 
Trigonometry we have 

p _ sin[^-0] % 

- ^-^sTn^TJir^]' ' • • • W 

in which & stands for a -f- 6, for brevity, being the angle 
which P makes with the vertical. N makes an angle = /? — 
with the vertical. 

The value, \ of 0, which makes P a maximum is found 

dP 

by placing - — — 0. From eq. (1), remembering that G is a 

function of 0, and that /3 and S are constants, we have 



sin (|B + 8 - $)["— sin 03 - <t>) - G cos (j8 - <J>)~| + Gsin (jB - </>) cos + 6-0) 
Ld<£ J 



-dG 

: 

d£ ~~ sin* [/3 + S - <l>] 



For P to be a maximum we must put 

numerator of above = (a) 



To find a geometrical equivalent of — — , denote A C by Z, 

and draw AE, making an angle = dcp with A C. Now the 
area ACT = AI X iOE=(Z + dZ)iZd0 = iZ 2 d<p . . . 
(neglecting infinitesimal of 2d order). Now 

,7/7 

dG = y X area J. (7/ X unity ; .*. -=— = i/Z 2 ; .*. (a) becomes 

acp 



sin (/? + S— fyiyZ* sin (/?— 0)- sin (/? + d — <p)G cos(/?— 0) 

+ £ sin (/S - 0) cos (/? + <? - 0) = ; 

i.e., G = 

j^Z 2 sin (/? — 0) sin (/? + 6 - 0) 

sin {/3 + S — 0) cos (/? — 0) — cos (/? + tf — 0) sin (/? — 0) 



576 MECHANICS OF ENGINEERING. 

when P is a maximum ; and hence, calling G' and <j>' and Z' 
the values of G, 0, and Z, for max. P, we have 

G' = ±yL n sm ({3-0') sin (/3 + S-cp') 

2 sin tf > • • \ / 

and therefore from (1) P max. itself is 

P' = ^yL-^^-^l (3) 

y sin d \ ' 

447. Geometric Interpretation and Construction. — If in Fig. 

496 we draw OF, making angle d with AD, C being any 
point on the ground surface BD, we have 

sin tf 
Drop a perpendicular ZZT from F to AG, and we shall have 

v y sin tf 

From this it follows that the weight of prism of base A CF 
and unit height 

= jyZ . TE= jyZ< ■ dn & ~ & S . iD ^ + S ~ »>. (4) 

sm d 

When J. 67 (as varies) assumes the position and value AC', 
bounding the prism of maximum thrust, Fig. 497, Z becomes 
— Z\ and = 0'; and eq. (4) gives the weight of the prism 
AC'F '. This weight is seen to be equal to that of the prism 
(or wedge) of maximum thrust ABC, by comparing eq. (4) 
with eq. (2); that is, AC' bisects the area ABC'F', and 
hence may be determined by fixing such a point C, on the 
upper profile BD, as to make the triangular area AC'F' 
equal to the sectional area of the wedge BC 'A; G'F' being 
drawn at an angle = d with AD. 

This holds for any form of ground surface BD, or any 



RETAINING WALLS. 



577 



values of the constants /?, a, or 6 
ally by trial, in dealing with 
an irregular profile BD. 

Having found AC, = 
£', P' can be found from 
(3), or graphically as fol- 
lows : (Fig. 497) With F' 
as a centre and radius = 
C'F', describe an arc cut- 
ting AD in J', and join 
C'J'. The weight of prism 
with base C'J'F' and unit height will 
has a weight 



C is best found graphic- 




Fig. 497. 



P'. For that prism 



= iy . F'J' 


. CM'; 


but 


F'J' = F'C 


and 


CM' = L' sir 



L' sin (0 - 0Q 
~~sin d ' 



weight of prism C'J'F = iyZ 



rT'j?-±„r>* 6in2 (P-<f>'). 



sin 6 



= P' 



[See eq. (3).] 



448. Point of Application of the Resultant Earth Thrust. — 
This thrust (called P' throughout this chapter except in the 
present paragraph) is now known in magnitude and direction, 
but not in position ; i.e., we must still determine its line of 
action, as follows : 

Divide AB into a number of equal parts, ab, be, cd, etc.; 
see Fig. 498. Treat ab as a small retaining wall, and find the 
magnitude P' of the thrust against it by § 447 ; treat ac simi- 
larly, thus finding the thrust, P" , against it ; then ad, ae, etc., 
the thrusts against them being found to be P" ' , P IV , etc. ; and 
so on. Now the pressure 

P' on ab is applied nearly at middle of ab, 



P" - P' 


a 


a 


u 


u 


be, 


pin pn 


a 


a 


u 


u 


cd, 




578 MECHANICS OF ENGINEEKING. 

and so on. Erect perpendiculars at the middle points of ab, 
be, cd, etc., equal respectively to P', 
P" - P', P" - P", etc., and join the 
ends of the perpendiculars. The per- 
pendicular through the centre of gravity 
of the area so formed (Fig. 498) will 
give, on AB, the required point of ap- 
plication of the thrust or earth pressure 
on AB, and this, with the direction and 
fig. 498. magnitude already found in § 447, will 

completely determine the thrust against the wall AB, 

449. Special Law of Loading. — If the material to be retained 
consists of loose stone, masses of masonry, buildings, or even 
moving loads, as in the case of a wharf or roadway, each can 
be replaced by the same weight of earth or other material 
which will render the bank homogeneous, situated on the same 
verticals, and the profile thus reduced can be treated by §§ 447 
and 448. 

Should the solid mass extend below the plane of rupture, 
AC, and the plane of natural slope, it will become a retaining 
wall for the material beyond, if strong enough to act as such 
(limiting the profile ABCD of Fig. 496 to the front of the 
mass, or to the front and line of rupture for maximum thrust 
above it, if it does not reach the surface); if not strong enough, 
or if it does not reach below the plane of natural slope, its 
presence is better ignored, probably, except that the increased 
weight must be considered. 

The spandrel wall of an arch may present two of these 
special cases ; i.e., the profile may be enlarged to include a 
moving load, while it may be limited at the back by the other 
spandrel. 

If the earth profile starts at the front edge of the top of 
wall, instead of from the back as at B, Fig, 496, eq. (3) would 
only apply to the portion behind AB prolonged, leaving the 
part on the wall (top) to be treated as a part of the wall to aid 
in resisting the thrust. 

If the wall is stepped in from the footings, or foundation 



RETAINING WALLS. 



579 



courses, probably the weak section will be just above them ; if 
stepped at intervals up the back of the wall, the surface of separa- 
tion between the wail and filling, if it is plane, will probably 
pass through the first step and incline forward as much as pos- 
sible without cutting the wall. 

450. Straight Earth-profile. — The general case can be simpli- 
fied as follows (the earth-profile BD being straight, at angle 
= C with vertical, = DET) : Since the triangles ABC and 




C'AF' are equal, from § 447, and AC is common, therefore 
BS=F'H (both being drawn ~| to AC). Draw AE and 
BM || to F'C (i.e., at angle d with AD), cutting DB, pro- 
longed, in E. We have 



BE 



EA 



CE EA - CF 



: . , and 



CE 



EA 



BE EA - BM 



But CF' = BM (since BS = H'F') ; 



therefore 



BE _ CE 
CE ~ BE 



i.e., BE . BE = CE\ 



which justifies the following construction for locating the de- 
sired point C on BD, and thus finding AC — L r and the 
angle 0': Describe a circle on ED as a diameter, and draw 



580 



MECHANICS OF ENGINEERING. 



BX 1 to jBD, thus fixing X in the curve. With centre E 
describe a circu lar arc through X, cutting BD in C, required. 

Having AC' (i.e., L'\ <p' is known ; hence from eq. (3) we 
obtain the earth thrust or pressure P'\ or, with F' as centre 
and radius = CF' , describe arc CJ'\ then the triangle G'F'J' 
is the base of a prism of unity height whose weight = P' (as 
in § 447). 

Centre of Pressure. — Applying the method of § 448, Fig. 
498, to this case, we find that the successive L' 's are propor- 
tional to the depths ab, ac, ad, etc., and that the successive i^'s 
are proportional [see (3)] to the squares of the depths ; hence 
the area in Fig. 498 must be triangular in this case, and the 
point of application of the resultant pressure on AB is one 
third of AB from A : just as with liquid pressure. 

451. Resistance of Retaining Walls. — (Fig. 500.) Knowing 
the height of the wall we can find its weight, = G 1 , for an as- 
sumed thickness, and unity width ~] to paper. The resultant 
of G 1 , acting through the centre of gravity of wall, and P' , the 
thrust of the embankment, in its proper 
line of action, should cut the base A V 
within the middle third and make an 
angle with the normal (to the base) less 
than the angle of friction. 

For the straight wall and straight 
earth-profile of Fig. 499 and § 450, the 
length Z', = AC, can be expressed in 
terms of the (vertical) height, h, of wall, thus : 




Fig. 500. 



AB = 



h 



cos a 



m& Z' = AC = AB 



sin (C — a) 



h 



sin (C — a) 



sin (C — 0') ~ " cos a ' sin (C — 00 ' 
eq. (3) becomes 

A 2 sin 2 (/? - 00 sin 8 (C - a) 



\r 



= iv 



jf 



cos 2 a sin d sin 2 (C — ) ' coi 

[A representing the large fraction for brevity.] 



.^.(5) 



fi 



RETAINING WALLS. 



581 



This equation will require, for a wall of rectangular section, 
that the thickness, d, increase as h, in order that its weight may 
increase as A 2 (i.e., as P') and that its resisting moment may 
increase with the overturning moment. 

By this equality of moments is meant that P'a = GJ) ; 
where a and h are the respective lever-arms of the two forces 
about the front edge of the middle third. {AB is the back of 
the wall.) In other words, their resultant will pass through 
this point. 

The following table is computed on the basis just mentioned, 
viz., that the resultant of P' and G shall jpass through the 
front edge of the midde third. 

The symbols of eq. (5) and the table are all shown in Fig. 
499, except y, 6, and d. y ■= weight of a cubic foot of earth, 
here assumed = § that of masonry (e.g., if earth weighs 
100 lbs., masonry is assumed to weigh 150 lbs. per cubic foot) ; 
6 = angle which the thrust P' makes with the back of the 
wall ; and d = a -\- 6, = 6 in this case as the wall is vertical, 
or a = 0. d is the proper safe thickness to be given to the 
wall, of rectangular section, to prevent overturning, as stated 
above ; h is the altitude, and A is the fraction shown in eq. (5). 

Whether the wall is safe against sliding on its base, and 
whether a safe compression per unit area is exceeded on the 
front edge of the base, are matters for separate consideration. 
The latter will seldom govern with ordinary retaining walls. 





a = 0; i.e., wall is vertical; also density of wall 


= f that of the earth. 






I. 


II. 


m. 








£ = 90° 






£ = 90 c 






C = /3 










e = 90° 






= /s 






6 = fi 




tan /3 


P 


p 


A 


d 


*' 


A 


d 


0' 


A 


d 


1.0 


45° 


22£° 


.17 


.Uh 


26° 


.18 


.22h 


45° 


.71 


.337* 


1.5 


56i° 


28° 


.29 


.447* 


33° 


.26 


.307* 


56° 


.83 


.437* 


2.0 


63+° 


31f° 


.38 


Mh 


38° 


.33 


.367* 


63° 


.89 


.517* 


4.0 


76° 


38° 


.61 


.647* 


45° 


.54 


.507* 


76° 


.97 


.657* 


Infinity 


90° 


45° 


1.00 


.827* 


90° 


1.00 


.827* 


90° 


1.00 


.827* 



In Case I of table, since a = 0, = 90° and C = 90° ; 
6 = 90°, and hence C'F' of Fig. 499 is "1 to AD, so that 



582 



MECHANICS OF ENGINEEEING. 



tan 2 £/?. . . . 


(6) 


1 d=fi, .\ d'= 


= fil 


sin 2 (/?-0') 


(7) 


sin /? cos 2 0' ' 



(since the area of aABC = aACF') 0' must = £/& 
These values, in (5), give 

P' = i r h 2 tan 3 J£ ; i.e., A 

In Case II, since C = 90°, a = and 6 
and (5) reduces to 

2K sin/? cos 2 0" ' 

In Case III, C = P and ^.Z? will be || to J.D, Z> being at 
^ j^ infinity. See Fig. 501. Through 
B draw .## ~\ to AD, and ^" 
making angle 6 with ^47). 6 r/ is 
^\^T >now to be located on BD. so as 
to make (area of) A ABC — 
(area of) aACF' (according 
to §447), the angle C'F'A being 
= d = a-\- 0\ =6, in this case, 
and hence also = j3. Conceive 
fig. 501. B and F' to be joined. 

Now &ACF' = aABF" + aBF'F". 

But A ABC = ABF'F" (equal bases and altitudes). 

Hence A ABC cannot = aACF' unless C is moved out 
to infinity ; and then 0' becomes = /?, and eq. (5) reduces to 




wBm 



P'— ^yh? sin /3 ; i.e., ^ = sin (3. 



(8) 



[Increasing a from zero will decrease the thickness d ; i.e., 
inclining the wall inwards will decrease the required thickness, 
but diminish the frictional stability at the base, unless the lat- 
ter be 1 to AB. The back of the wall is frequently inclined 
outwards, making the section a trapezoid, to increase the fric- 
tional stability at the base when necessary, as with timber 
walls supporting water.] 



EETAINING WALLS. 583 

452. Practical Considerations. — An examination of the 
values of A and d in the table of § 451 will show that in sup- 
porting quicksand and many kinds of clay which are almost 
fluid under the influence of water, it is important to know 
what kind of drainage can be secured, for on that will depend 
the thickness of the wall. With well compacted material free 
from water-bearing strata, an assumed natural slope of 1J- to 1 
(i.e., li hor. to 1 vert.) will be safe ; the actual pressure below 
the effect of frost and surface water will be that due to a much 
steeper slope on account of cohesion (neglected in this theory). 

The thrust from freshly placed material can be reduced by 
depositing it in layers sloping back from the wall. If it is not 
so placed, however, the natural slope will seldom be flatter 
than 1J to 1 unless reduced by water. In supporting material 
which contains water-bearing strata sloping toward the wall 
and overlain by strata which are liable to become semi-fluid 
and slippery, the thrust may exceed that due to semi-fluid ma- 
terial on account of the surcharge. If these strata are under 
the wall and cannot be reached by the foundation, or if resist- 
ance to sliding cannot be obtained from the material in front 
by sheet-piling, no amount of masonry can give security. 

Water at the back of the wall will, by freezing, cause the 
material to exert an indefinitely great pressure, besides disinte- 
grating the wall itself. If there is danger of its accumulation, 
drainage should be provided by a layer of loose stone at the 
back leading to " weep-holes" through the wall. 

A friction-angle at the back of the wall equal to that of the 
filling should always be realized by making the back rough by 
steps, or projecting stones or bricks. Its effect on the required 
thickness is too great to be economically ignored. 

The resistance to slipping at the base can be increased, when 
necessary, by inclining the foundation inwards; by stepping 
or sloping the back of the wall so as to add to its effective 
weight or incline the thrust more nearly to the vertical; by 
sheet-piling in front of the foundation, thus gaining the resist- 
ance offered by the piles to lateral motion ; by deeper founda- 
tions, gaining the resistance of the earth in front of the wall. 



584 MECHANICS OF ENGINEERING. 

The coefficient of friction on the base ranges, according to 
Trau twine, from 0.20 to 0.30 on wet claj ; 
" .50 to .66 " dry earth ; 
" .66 to .75 " sand or gravel ; 
" .60 on a dry wooden platform ; to .75 on a 
wet one. 

If the wall is partially submerged, the buoyant effort should 
be subtracted from G 1 , the weight of wall. 

453. Results of Experience. — (Trautwine.) In railroad prac- 
tice, a vertical wall of rectangular section, sustaining sand, 
gravel, or earth, level with the top [p. 682 of Civ. Eng. Pocket 
Book] and loosely deposited, as when dumped from carts, cars, 
etc., should have a thickness d, as follows : 

If of cut stone, or of first-class large ranged rubble, in mortar. , . d = . 35h 

" good common scabbled mortar-rubble, or brick , .d — AQh 

" well scabbled dry rubble d= .50h 

"Where h includes the total height, or about 3 ft. of foundations. 

(a) For the best masonry of its class h may be taken from 
the top of the foundation in front. 

(b) A mixture of sand or earth, with a large proportion of 
round boulders or cobbles, will weigh more than the backing 
assumed above ; requiring d to be increased from one eighth to 
one sixth part. 

(e) The wall will be stronger by inclining the back inwards, 
especially if of dry masonry, or if the backing is put in place 
before the mortar has set. 

(d) The back of the wall should be left rough to increase 
friction. 

. (e) Where deep freezing occurs, the back should slope out- 
ward for 3 or 4 feet below the top and be left smooth. 

(f) When a wall is too thin, it will generally fail by bulging 
outward at about one third the height. The failure is usually 
gradual and may take years. 

(g) Counterforts, or buttresses at the back of the wall, usually 
of rectangular section, may be regarded as a waste of ma- 
sonry, although considerably used in Europe; the bond will 



KETAINING WALLS. 585 

seldom hold them to the wall. Buttresses in front add to the 
strength, but are not common, on account of expense. 

(h) Land-ties of iron or wood, tying the wall to anchors im- 
bedded below the line of natural slope, are sometimes used to 
increase stability. 

(i) Walls with curved cross-sections are not recommended. 

454. Conclusions of Mr. B. Baker. — (" Actual Lateral Pressure 
of Earthwork.") Experience has shown that d — 0.25A, with 
batter of 1 to 2 inches per foot on face, is sufficient when 
backing and foundation are both favorable ; also that under no 
ordinary conditions of surcharge or heavy backing, with solid 
foundation, is it necessary for d to be greater than 0.50A. 

Mr. Baker's own rule is to make d = 0.33A at the top of 
the footings, with a face batter of 1J inches per foot, in ground 
of average character ; and, if any material is taken out to form 
a face-panel, three fourths of it is put back in the form of a 
pilaster. The object of the batter, and of the panel if used, is 
to distribute the pressure better on the foundation. All the 
walls of the " District Railway" (London) were designed on 
this basis, and there has not been a single instance of settle- 
ment, of overturning, or of sliding forward. 

455. Experiments with Models. — Accounts of experiments 
with apparatus on a small scale, with sand, etc., may be found 
in vol. lxxi of Proceedings of Institution of Civil Engineers, 
London, England (p. 350) ; also in vol. n of the " Annales des 
Ponts et Chaussees" for 1885 (p. 788). 

The results of these experiments, and the results of experi- 
ence given in §§ 453 and 454, when compared with the table 
of p. 581, indicate a fairly close agreement between practice 
and theory. This agreement is believed to be close enough 
so that the general method of §§ 447 and 451, with the table 
of p. 581, can be relied upon in practice. The greatest value of 
this method will, of course, be for cases of exceptional loading, 
inclined walls, etc., where the results of experience do not 
furnish so valuable a guide. 



CHAPTEE IT. 

HYDROSTATICS ((hntinued)— IMMERSION AND FLOTATION. 

456. Rigid Body Immersed in a Liquid. Buoyant Effort. — If 
any portion of a body of homogeneous liquid at rest be con- 
ceived to become rigid without alteration of shape or bulk, it 
would evidently still remain at rest ; i.e., its weight, applied at 
its centre of gravity, would be balanced by the pressures, ou its 
bounding surfaces, of the contiguous portions of the liquid ; 
hence, 

If a rigid body or solid is immersed in a liquid, both being 
at rest, the resultant action upon it of the surrounding liquid 
{or fluid) is a vertical upward force called the "buoyant 
effort" equal in amount to the weight of liquid displaced, 
and acting through the centre of gravity of the volume {con- 
sidered as homogeneous} of displacement {now occupied by the 
solid). This point is called the centre of buoyancy, and is 
sometimes spoken of as the centre of gravity of the displaced 
water. If V = the volume of displacement, and y = heavi- 
ness of the liquid, then the 

buoyant effort = V'y (1) 

(By " volume of displacement" is meant, of course, the volume 
of liquid actually displaced when the body is immersed.) 

If the weight G' of the solid is not equal to the buoyant 
effort, or if its centre of gravity does not lie in the same verti- 
cal as the centre of buoyancy, the two forces form an unbal- 
anced system and motion begins. But as a consequence of 
this very motion the action of the liquid is modified in a man- 
ner dependent on the shape and kind of motion of the body. 

586 



IMMERSION. 



587 



Problems in this chapter are restricted to cases of rest, i.e., 
balanced forces. 

Suppose G' = V'y ; then, 

If the centre of gravity lies in the same vertical line as the 
centre of buoyancy and underneath the latter, the equilibrium 
is stable / i.e., after a slight angular disturbance the body re- 
turns to its original position (after several oscillations); while 
if above the latter, the equilibrium is unstable. If they coin- 
cide, as when the solid is homogeneous (but not hollow), and 
of the same heaviness (§ 7) as the liquid, the equilibrium is 
indifferent, i.e., possible in any position of the body. 

The following is interesting in this connection : 

In an account of the new British submarine boat " Nautilus," 
a writer in Chambers's Journal remarked [1887] : "At each 
side of the vessel are four port-holes, into which fit cylinders 
two feet in diameter. When these cylinders are projected 
outwards, as they can be by suitable gearing, the displacement 
of the boat is so much increased that the vessel rises to the 
surface; but when the cylinders are withdrawn into their 
sockets, it will sink." 

As another case in point, large water-tight canvas " air-bags" 
have recently been used for raising sunken ships. They are 
sunk in a collapsed state, attached by divers to the submerged 
vessel, and then inflated with air from pumps above, which of 
course largely augments their displacement while adding no 
appreciable weight. 



457. Examples of Immersion. — 
ample of stable equi- 
librium, the centre of 
buoyancy B being above 
the centre of gravity C, 
and the buoyant effort 

V'y = G' = the weight 
of the solid ; at (a'), con- 
versely, we have un- 
stable equilibrium, with 

V'y still = G'. At (b) the buoyant effort V'y is > G\ and 




Fig. 502. 



588 MECHANICS OF ENGINEERING. 

to preserve equilibrium the body is attached by a cord to the 
bottom of the vessel. The tension in this cord is 

s b = V'y - & (i) 

At (c) V'y is < G', and the cord must be attached to a 
support above, and its tension is 

S e = (?' - V'y (2) 

If in eq. (2) [(<?) in figure] we call S c the apparent weight of 
the immersed body, and measure it by a spring- or beam-bal- 
ance, we may say that 

The apparent weight of a solid totally immersed in a liquid 
equals its real weight diminished by that of the amount of 
liquid displaced / in other words, the loss of weight = the 
weight of displaced liquid. 

Example 1. — How great a mass (not hollow) of cast-iron can 
be supported in water by a wrought-iron cylinder weighing 
140 lbs., if ^he latter contains a vacuous space and displaces 

3 cub. feet of water, both bodies being completely immersed ? 
[Ft., lb., sec] 

The buoyant effort on the cylinder is 

V'y = 3 X 62.5 = 187.5 lbs., 

leaving a residue of 47.5 lbs. upward force to buoy the cast- 
iron, whose volume V" is unknown, while its heaviness (§ 7) 
is y" = 450 lbs. per cub. foot. The direct buoyant effort of 
the water on the cast-iron is V"y = \_V" X 62.5] lbs., 
and the problem requires that this force -f- 47.5 lbs. shall 
= V'y" = the weight G" of the cast-iron ; 

.-. V" X 62.5 + 47.5 = V" X 450 ; 

.-. V" = 0.12 cub. ft., while 0.12 X 450= 54 lbs. of cast-iron. 

Ans. 
Example 2. — Required the volume V\ and heaviness y\ 
of a homogeneous solid which weighs 6 lbs. out of water and 

4 lbs. when immersed {apparent weight) (ft., lb., sec ). 



IMMERSION. 589 

From eq. (2), 4 = 6 - V X 62.5 ; .\ V = 0.032 cub. feet ; 

.-. y> = G f + V' = 6 -?- 0.032 = 187.5 lbs. per cub. ft., 

aud the ratio of 7/ to y is 187.5 : 62.5 = 3.0 (abstract num- 
ber) ; i.e., the substance of this solid is three times as dense, 
or three times as heavy, as water. [The buoyant effort of the 
air has been neglected in giving the true weight as 6 lbs.] 

458. Specific Gravity. — By specific gravity is meant the ratio 
of the heaviness of a given homogeneous substance to that of 
a standard homogeneous substance ; in other words, the ratio 
of the weight of a certain volume of the substance to the 
weight of an equal volume of the standard substance. Dis- 
tilled water at the temperature of maximum density (4° Centi- 
grade) under a pressure of 14.7 lbs. per sq. inch is sometimes 
taken as the standard substance, more frequently, however, at 
62° Fahrenheit (16°.6 Centigrade). Water, then, being the 
standard substance, the numerical example last given illustrates 
a common method of determining experimentally the specific 
gravity of a homogeneous solid substance, the value there ob- 
tained being 3. The symbol a will be used to denote specific 
gravity, which is evidently an abstract number. The standard 
substance should always be mentioned, and its heaviness;/; 
then the heaviness of a substance whose specific gravity is <x is 

r' = <ry, (1) 

and the weight G' of any volume V of the substance may be 
written 

G' = Vy = Very (2) 

Evidently a knowledge of the value of y' dispenses with the 
use of <r, though when the latter can be introduced into prob- 
lems involving the buoyant effort of a liquid the criterion as 
to whether a homogeneous solid will sink or rise, when im- 
mersed in the standard liquid, is more easily applied, thus : 
Being immersed, the volume V of the body = that, V, of 
displaced liquid. Hence, 



590 



MECHANICS OF ENGINEEKING. 



if G' is > V'y, i.e., if V'y' is > V'y, or or > 1, it sinks ; 
while if G' is < V'y, or a < 1, it rises ; 

i.e., according as the weight G' is > or < than the buoyant 
effort. 

Other methods of determining the specific gravity of solids, 
liquids, and gases are given in works on Physics. 

459. Equilibrium of Flotation. — In case the weight G' of an 
immersed solid is less than the buoyant effort V'y (where V is 
the volume of displacement, and /the heaviness of liquid) the 
body rises to the surface, and after a series of oscillations comes 
to rest in such a position, Fig. 503, that its centre of gravity C 
and the centre of buoyancy B (the new B, belonging to the 
new volume of displacement, which is limited above by the 
horizontal plane of the free surface of the liquid) are in the 
same vertical (called the axis of flotation, or line of support), 
and that the volume of displacement has diminished to such a 
new value V, that 

Vy=G' (1) 

In the figure, V= vol. AND, below the horizontal plane 
AN, and the slightest motion of the body will change the form 
e of this volume, in general (whereas with 

complete immersion the volume of dis- 
placement remains constant). For stable 
equilibrium it is not essential in every 
case that (7 (centre of gravity of body) 
should be below B (the centre of buoy- 
ancy) as with complete immersion, since if 
the solid is turned, B may change its posi- 
tion in the body, as the form of the volume AND changes. 

There is now no definite relation between the volume of 
displacement Fand that of the body, V , unless the latter is 
homogeneous, and then for G' we may write V'y', i.e. 

V'y' = Vy (for a homogeneous solid) ; . 




Fig. 503. 



(2) 



or, the volumes are inversely proportional to the heavinesses. 



FLOTATION. 



591 



The buoyant effort of the air on the portion ANE 'may be 
neglected in most practical cases, as being insignificant. 

If the solid is hollow, the position of its centre of gravity C 
may be easily varied (by shifting ballast, e.g.) within certain 
limits, but that of the centre of buoyancy B depends only on 
the geometrical form of the volume of displacement AND, 
below the horizontal plane AN. 

Example. — (Ft., lb., sec.) Will a solid weighing G' = 400 
lbs., and having a volume V = 8 cub. feet, without hollows 
or recesses, float in water? To obtain a buoyant effort of 
400 lbs., we need a volume of displacement, see eq. (1), of 



V= 



G' _ 400 
T ~~ 62.5 



only 6.4 cub. ft. 



Hence the solid will float with 8 — 6.4, or 1. 6, cub. ft. pro- 
jecting above the water level. 

Query : A vessel contains water, reaching to its brim, and 
also a piece of ice which floats without touching the vessel. 
When the ice melts will the water overflow? 

460. The Hydrometer is a floating instrument for determin- 
ing the relative heavinesses of liquids. Fig. 504 shows a sim- 
ple form, consisting of a bulb and a cylin- 
drical stem of glass, so designed and 
weighted as to float upright in all liquids 
whose heavinesses it is to compare. Let F 
denote the uniform sectional area of the 
stem (a circle), and suppose that when float- 
ing in water (whose heaviness = y) the 
water surface marks a point A on the stem ; 
and that when floating in another liquid, 
say petroleum, whose heaviness, =^ p , we 
wish to determine, it floats at a greater 
depth, the liquid surface now marking A' 
on the stem, a height — x above A. G' is 
the same in both experiments; but while the volume of dis- 
placement in water is V, in petroleum it is V "+ Fx. There- 
fore from eq. (1), § 459, 




Fig. 504. 



592 MECHANICS OF ENGINEERING. 

in the 'water G' = Vy, (1) 

and in the petroleum G' = ( V-\-Fx)y p ; . . (2). 

from which, knowing G\ F, x, and y, we find Fand y p , i.e., 

r =f* r ' = vife- • • • (3) 

[K\B. — i^is best determined by noting the additional dis- 
tance, = I, through which the instrument sinks in water under 
an additional load P, not immersed / for then 

G'+p=(r+Fi) r , or >=-£] 

Example. — [Using the inch, ounce, and second, in which 
system y = 1000 + 1728 = 0.578 (§ 409).] With G' = 3 
ounces, and i^= 0.10 sq. inch, x being observed, on the 
graduated stem, to be 5 inches, we have for the petroleum 



r > = 3 + 0.10X5XQ.578 = °^ ° Z ' ^ CUblC 1DCh 

== 56.7 lbs. per cub. foot. 



Temperature influences the heaviness of most liquids to- 
some extent. 

In another kind of instrument a scale-pan is fixed to the top 
of the stem, and the specific gravity computed from the weight 
necessary to be placed on this pan to cause the hydrometer to 
sink to the same point in all liquids for which it is used. 

461. Depth of Flotation. — If the weight and external shape 
of the floating body are known, and the centre of gravity so 
situated that the position of flotation is known, the depth of 

the lowest point below the surface may be determined. 



FLOTATION. 



593 



Case I. Right prism or cylinder with its axis vertical. — 
Fig. 505. (For stability in this position, 
see § 464(2.) Let G' — weight of cylin- 
der, Fthe, area of its cross-section (full -^p? 
circle), hf its. altitude, and h the un- HH 
known depth of flotation (or draught) ; -^ 
then from eq. (1), § 426, 

G' = Fh Y] /. h=^\. (1) 



rhh 




Fig. 505. 



in which y = heaviness of the liquid. 
If the prism (or cylinder) is homo- 
geneous (and then C, at the middle of h\ is higher than B) 
and y' its heaviness, we then have' 



*=*»£=*£ = ,*' 



Fy 



y 



(2) 



in which <j = specific gravity of solid referred to the liquid as 
standard. (See § 458.) 

Case II. Pyramid or cone with axis vertical and vertex 
down. — Fig. 506. Let V — volume of 
whole pyramid (or cone), and V= vol- 
ume of displacement. From similar 
pyramids, 







.h'. 



G 



But G' = Vy ; or, V = — ; whence 



-|ggr 


§§3§!1 


HJ^-I^\ \f3^EJ=^L 




^zr^-SEEE^yf.JFr^^A=. 



Fig. 506. 



h = h 



, s n? 



(3) 



594 



MECHANICS OF ENGINEERING. 



Case III. Ditto, 



\h" 





1 

1 

__) 


\ - A= 








■>*— . 


— =— 


•"^^ : 



ut vertex up. — Fig. 507. Let the nota- 
tion be as before, for V and V. The 
part out of water is a pyramid of volume 
= V" — V — 7, and is similar to the 
whole pyramid ; 

... v f - V: V :: A //3 : h'\ 
G r 



Also, 



V 



Fig. 507. 



(4) 



/., finally, A = A'[l - j/l - [G' -r- V'y]\ . . ■ 

Case IY. Sphere. — Fig. 508. The volume immersed is 
V = f\nx*)dz = 7tf\%rz - z 2 )dz = nltXr - |1 ; 

and hence, since Vy = G' = weight 
of sphere, 

7t rh>-^ = ^. . . (5) 

3 y 

-iv^IfII From which cubic equation h may be 
i^ ^S^£ ^^p|f|Z obtained by successive trials and ap- 

f=|f^ J =: ^-~ = proximations. 

fig- 508. [An exact solution of (5) for the 

unknown h is impossible, as it falls under the irreducible case 
of Cardan's Rule.] 

Case Y. Might cylinder with axis horizontal. — Fig. 509. 

mers. Jv\ = t area of ^ ADB \ * l 

= (r*a — %r* sin 2a)l ; 
G' 





-=r hence, since V ' = 

1^7509. Y 

c 

Vr\a — ■§- sin 2a] = — , 
Y 



(6) 



EXAMPLES OF FLOTATION. 595 

From this transcendental equation we can obtain a, by trial, 
in radians (see example in § 428), and finally A, since 

h = r(l — cos a) (7) 

Example 1. — A sphere of 40 inches diameter is observed to 
have a depth of flotation h = 9 in. in water. Required its 
weight G' . From eq. (5) (inch, lb., sec.) we have 

G' = [62.5 4- 1728]tt9 2 [20 - ± x 9] = 156.5 lbs. 

The sphere may be hollow, e.g., of sheet metal loaded with 
shot ; constructed in any way, so long as G' and the volume 
Voi displacement remain unchanged. But if the sphere is 
homogeneous, its heaviness (§ 7) y' must be 

= G' -v- V = G' -T- f Ttr 3 = (156.5) -^ f7r20 3 
= .00466 lbs. per cubic inch, 

and hence, referred to water, its specific gravity is cr = about 
0.13. 

Example 2. — The right cylinder in Fig. 509 is homogeneous 
and 10 inches in diameter, and has a specific gravity (referred 
to water) of cr = 0.30. Required the depth of flotation h. 

Its heaviness must be y' = cry ; hence its weight 

G' = Very = TzrHcry ; 
hence, from eq. (6), 

r*l[a — \ sin 2a] = nrHcr^ ,\ a — \ sin 2a = ncr 

(involving abstract numbers only). Trying a = 60° ( = \n in 
radians), we have 

\n — \ sin 120° =± 0.614 ; whereas ncr = .9424 

For a = 70°, 1.2217 - J sin 140° = 0.9003 ; 
For a = 71°, 1.2391 - \ sin 142° = 0.9313 ; 
For a = 71° 22 / , 1.2455 - i sin 142° 44' = 0.9428, which may 
be considered sufficiently close. Now from eq. (7), 

h = (5 in.) (1 - cos 71° 22') = 3.40 m.—Ans. 



596 MECHANICS OF ENGINEERING. 

462. Draught of Ships. — In designing a ship, especially if ot 
a new model, the position of the centre of gravity is found by 
eq. (3) of § 23 (with weights instead of volumes) ; i.e., the sum 
of the products obtained by multiplying the weight of each 
portion of the hull and cargo by the distance of its centre of 
gravity from a convenient reference-plane (e.g., the horizontal 
plane of the keel bottom) is divided by the sum of the weights, 
and the quotient is the distance of the centre of gravity of the 
whole from the reference-plane. 

Similarly, the distance from another reference-plane is de- 
termined. These two co-ordinates and the fact that the centre 
of gravity lies in the median vertical plane of symmetry of the 
ship (assuming a symmetrical arrangement of the framework 
^nd cargo) fix its location. The total weight, G' , equals, of 
course, the sum of the individual weights just mentioned. The 
■ventre of buoyancy, for any assumed draught and correspond- 
ing position of ship, is found by the same method ; but more 
simply, since it is the centre of gravity of the imaginary homo- 
geneous volume between the water-line plane and the wetted 
surface of the hull. This volume (of "displacement") is 
divided into an even number (say 4 to 8) of horizontal laminae 
of equal thickness, and Simpson's Rule applied to find the vol- 
ume (i.e., the V of preceding formulae), and also (eq. 3, § 23) 
the height of its centre of gravity above the keel. Similarly, 
by division into (from 8 to 20) vertical slices, ~[ to keel (an 
even number and of equal thickness), we find the distance of 
the centre of gravity from the bow. Thus the centre of buoy- 
ancy is fixed, and the corresponding buoyant effort Yy (tech- 
nically called the displacement and usually expressed in tons) 
computed, for any assumed draught of ship (upright). That 
position in which the " displacement" = G' = weight of ship 
is the position of equilibrium of the ship when floating up- 
right in still water, and the corresponding draught is noted. 
As to whether this equilibrium is stable or unstable, the fol- 
lowing will show. 

In most ships the centre of gravity C is several feet above 
the centre of buoyancy, B, and a foot or more below the water 
line. 



DKAUGHT OF SHIPS. 



597 



After a ship is afloat and its draught actually noted its total 
weight G\ = Vy, can be computed, the values of Ffor dif- 
ferent draughts having been calculated in advance. In this 
way the weights of different cargoes can also be measured. 

Example.— A ship having a displacement of 5000 tons is 
itself 5000 tons in weight, and displaces a volume of salt water 
V= G'+ y = 10,000,000 lbs. -j- 64 lbs. per cub. ft. = 156250 
cub. ft. 



463. Angular Stability of Ships. — If a vessel floating upright 
were of the peculiar form and position of 
Fig. 510 (the water-line section having an 
area = zero) its tendency to regain that 
position, or depart from it, when slightly 
inclined an angle from the vertical is due 
to the action of the couple now formed by 
the equal and parallel forces Vy and G\ 
which are no longer directly opposed. This 
couple is called a righting couple if it acts 
to restore the first position (as in Fig. 511, 
where C is lower than B), and an 
upsetting couple if the reverse, C 
above B. In either case the mo- 
ment of the couple is 

= Vy . BG sin = Vye sin 0, ;r= 




Fig. 510. 




Fig. 511. 



and the centre of buoyancy B does not 
change its position in the vessel, since 
the water-displacing shape remains 
the same ; i.e., no new portions of 
the vessel are either immersed or 
raised out of the water. 

But in a vessel of ordinary form, when turned an angle 0from 
the vertical, Fig. 512 (in which ED is a line which is vertical 
when the ship is upright), there is a new centre of buoyancy, 
B x , corresponding to the new shape A y NJ) of the displacement- 
volume, and the couple to right the vessel (or the reverse) 



MECHANICS OF ENGINEERING. 



consists of the two forces G' at C and Vy at B l , and has a 

moment (which we may call Jf/, or 
'e moment of stability) of a value 

(§28) 




Vy . m6 7 sin 0. 



(i) 



n _jz_— Now conceive put in at B (centre 
2= of buoyancy of the upright posi- 
^ tion) two vertical and opposite 
— ■ forces, each = Vy = G f , calling 
them P and P x (see § 20), Fig. 512. 
We can now regard the couple \_G\ Vy] as replaced by the 
two couples [G\ P] and [P 15 Vy] ; for evidently 



Fig. 512. 



Vy . mC sin = Vy , BC sin -f~ "^V • m ^ sm 5 
(§§33 and 34;) 

.-. M=Vy BC sin + Vy ln~B sin 0. . . (2) 

But the couple [G\ P~\ would be the only one to right the 
vessel if no new portions of the hull entered the water or 
emerged from it, in the inclined position ; hence the other 
couple \P XJ Vy] owes its existence to the emersion of the 

wedge AOA iy and the immersion 
of the wedge NON x \ i.e., to the 
loss of a buoyant force Q = (vol- 
ume A OA x ) X y on one side, and the 
4— gain of an equal buoyant force on 
the other; therefore this couple 
[jP 1? Vy] is the equivalent of the 
- couple [Q, Q], Fig. 51L, formed by 
putting in at the centre of buoyancy 
of each of the two wedges a vertical 
force 




Q = (vol. of wedge) X v — V w y. (See figure.) 



STABILITY OF SHIPS. 599 

If a denotes the arm of this couple, we may write 



Vy . m£ sin 0, [of eq. (2)], = V w ya ; . . (3) 

and hence, denoting BC by e, we have 

M = ± Vye sin + V w ya ; .... (4) 

the negative sign in which is to be used when C is above B 
(as with most ships). (2, the intersection of ED and AJV, 
does not necessarily lie on the new water-line plane A l JV 1 . 

Example. — If a ship of ( Vy =) 3000 tons displacement 
with C 4 ft. above B (i.e., e = — 4 ft.) is deviated 10° from 
the vertical, in salt water, for which angle the wedges A OA x and 
NON x have each a volume of 4000 cubic feet, while the hori- 
zontal distance a between their centres of buoyancy is 18 feet, 
the moment of the acting couple will be, from eq. (4) (ft.-ton- 
sec. system, in which y of salt water = 0.032), 

M = - 3000 X 4 X 0.1736 + 4000 X 0.032 X 18 = 220.8 ft. tons, 

which being -j- indicates a righting couple. 

464. Remark. — If with a given ship and cargo this moment 
of stability, M, be computed, by eq. (4), for a number of values 
of 0, and the results plotted as ordinates (to scale) of a curve, 
being the abscissa, the curve ob- 
tained is indicative of the general 
stability of the ship. See Fig. 514. 
For some value of = OK(sls well 
as for = 0) the value of M is 
zero, and for > OK, M is nega- 
tive, indicating an upsetting couple. FlG - 514 - 
That is, for 0=0 the equilibrium is stable, but for = OK, 
unstable ; and M = in both positions. From eq. (4) we see 
why, if C is above B, instability does not necessarily follow. 

464a. Metacentre of a Ship. — Referring again to Fig. 512, 
we note that the entire couple [G\ Vy] will be a righting 
couple, or an upsetting couple, according as the point m (the 




600 



MECHANICS OF ENGINEEKING. 



intersection of the vertical through B 1 , the new centre of 
buoyancy, with BC prolonged) is above or below the centre 
of gravity C of the ship. The location of this point m changes 
with ; but as becomes very small (and ultimately zero) m 
approaches a definite position on the line DE, though not oc- 
cupying it exactly till = 0. This limiting position of m is 
called the metacentre, and accordingly the following may be 
stated : A ship floating upright is in stable equilibrium if its 
metacentre is above its centre of gravity / and vice versd. 
In other words, for a slight inclination from the vertical a 
righting, and not an upsetting, couple is called into action, if 
m is above C. To find the metacentre, by means of the dis- 
tance Bm, we have, from eq. (3), 



mB 



Vy sin 0' 



(5) 



and wish ultimately to make = 0. Now the moment 
( VwV)® — the sum of the moments about the horizontal fore- 
and-aft water-line axis OB, Fig. 515, of the buoyant efforts 



qjzdFy 




O^T 



due to the immersion of the 
separate vertical elementary 
prisms of the wedge OLNJL, 
plus the moments of those lost, 
from emersion, in the wedge 
OLA.A. Let OA.LN, be the 
new water-line section of the 
ship when inclined a small 
angle from the vertical 
old water-line. Let z = the 



Fig. 515. 

(0 = NO,N^ and OALN the 
"1 distance of any elementary area dF oi the water-line section 
from OL (which is the intersection of the two water-line 
planes). Each dF\% the base of an elementary prism, with 
altitude = 03, of the wedge N x OLN (or of wedge AfiLA 
when z is negative). The buoyant effort of this prism = (its 
vol.) X y = yzcpdF, and its moment about OL is (pyz'dF, 
Hence the total moment, t= Qa, or V w ya, of Fig. 513, 



<Pyfz\lF= ycf> X Iol 



THE METACENTRE. 601 

•of water-line section, in which I 0L denotes the " moment of 
inertia" (§ 85) of the plane figure OjlZJVO about the axis OL. 
Hence from (5), putting = sin (true when = 0), we have 
inB = I 0L -i- V\ and therefore the distance mC\ of the meta- 
centre m above C, the centre of gravity of the ship, Fig. 512, is 

^C, = h m , = 7 <* (° f water-lipe sec.) ± ^ > _ < (g) 

in w T hich e = BC — distance from the centre of gravity to the 
centre of buoyancy, the negative sign being used when (7 is 
above B\ while F= whole volume of water displaced by the 
ship. 

We may also write, from eqs. (6) and (1), for small values 

Mom. of righting cowple = M= Vy sin -~ ± e L . (7) 



or 



M=y sin <p[I 0L ± Ve~]. . . . . (7)' 



Eqs. (7) and (7)' will give close approximations for < 10° or 
15° with ships of ordinary forms. 

Example 1. — A homogeneous right parallelepiped, of 
heaviness y\ floats upright as in 

Fig. 516. Find the distance ^ \ ~* 

mC — h m for its metacentre in this A p- 1 
position, and whether the equilibrium ^= 
is stable. Here the centre of gravity, f-fj 
C, being the centre of figure, is of ~~ } 
course above B, the centre of buoy- 1? 
ancy ; hence e is negative. B is the £L 
centre of gravity of the displacement, =^ -~ - ~ — ~ 

and is therefore a distance %h below FlG ' 516 * 

the water-line. We here assume that I is greater than b'. 
From eq. (2), § 461, 

r 



602 MECHANICS OP ENGINEEKING. 

and since CD = \Ji\ and BD — %h, .\ e — £(A'~ h)\ 



i.e., e 



while (§ 90) I 0L , of the water-line section AJV, = -£%l'b n * 
Also, 

V=VKL' = VVh' ^ ; 

r 

and hence, from eq. (6), we have 

h __Wy_ _ iA ,r 1 _r / ]_ r r^-e^Yi - £'Y1 
^""lamy 2 L rJ~iMyL r^ r'-l 

Hence if Z> /2 is > 6A /a £ (l - K \ the position in Fig. 516 is 

one of stable equilibrium, and vice versa. E.g., if y' = \y y 
V = 12 inches and h! = 6 inches, we have (inch, pound, sec.) 



mi 



C = i[lM-6x¥(l - i)J = 2.5 in. 



The equilibrium will be unstable if, with y' = \y, V is made 
less than 1.225 h! ; for, putting m(7 = 0, we obtain 2/ = 
1.225 h'. 

Example 2. — (Ft., lb., sec.) Let Fig. 517 represent the half 
water-line section of a loaded ship of G' = Vy = 1010 tons 




Fig. 517. 



displacement ; required the height of the metacentre above the 
centre of buoyancy, i.e., inB = % (See equation just before eq. 
(6).) Now the quantity I OL , of the water-line section, may, 
from symmetry, (see § 93,) be written 

loL = zf\y i dx, (1) 



META CENTRE. 60S 

in which y = the ordinate ~\ to the axis OL at any point ; and 
this, again, by Simpson's Kule for approximate integration, 
OL being divided into an even nnmber, n, of equal parts, and 
ordinates erected (see figure), may be written 

Iol = |. ^"-[2/o 8 + %i 3 + 2/s 8 + • • • +yi-d 

+ % 2 3 + 2//+---+2/ 8 l - 2 ) + 2/ s „]- 

From which, by numerical substitution (see figure for dimen- 
sions ; n = 8), 

IoL = I * sS [ (0,53 + 4(53 + 129 + 183 + T) 

+ 2 (9 3 + 14 3 + ll 3 ) + 0.5 3 ]; 

125 

1728 729 

or > 2197 2744 

343 1331 



I 0L = 4A[0.125 +4 X 4393 + 2 X 4804+0.125] 

Inr. 120801 



= 120801 biquad. ft. ;,. m B = -f = ^^ + 
= 3.8 feet. 

That is, the metacentre is 3.8 feet above the centre of buoyancy, 
and hence, if i?C:=2 feet, is 1.90 ft. above the centre of 
gravity. [See Johnson's Cyclopaedia, article JVaval Architec- 
ture.'] 

465. Metacentre for Longitudinal Stability. — If we consider 
the stability of a vessel with respect to pitching, in a manner 
similar to that just pursued for rolling, we derive the position 
of the metacentre for pitching or for longitudinal stability — 
and this of course occupies a much higher position than that 
for rolling, involving as it does the moment of inertia of the 
water-line section about a horizontal gravity axis "| to the keel. 
With this one change, eq. (6) holds for this case also. In 
large ships the height of this metacentre above the centre of 
gravity of the ship may be as great as 90 feet. 



CHAPTER V. 



HYDROSTATICS {Continued)— GASEOUS FLUIDS. 




A 

2S 



|l I II I lYTTTT 



Fig. 518. 



466. Thermometers. — The temperature, or " hotness" of 
liquids has, within certain limits, but little influence on their 
statical behavior, but with gases must always be taken into 
account, since the three quantities, tension, temperature, and 
volume, of a given mass of gas are connected by a nearly in- 
variable law, as will be seen. 

An air-thermometer, Fig. 518, consists of a large glass bulb 
filled with air, from which projects a fine straight tube of 

even bore (so that equal lengths 
: ;'i represent equal volumes). A 
ti?5 small drop of liquid, A, sepa- 
'.;/ rates the internal from the ex- 
>s ternal air, both of which are 
at a tension of (say) one at- 
mosphere (14.7 lbs. per sq. inch). When the bulb is placed 
in melting ice (freezing-point) the drop stands at some point F 
in the tube; when in boiling water (boiling under a pressure 
of one atmosphere), the drop is found at B, on account of the 
expansion of the internal air under the influence of the heat 
imparted to it. (The glass also expands, but only about y^-g- 
as much ; this will be neglected.) The distance FB along the 
tube may now be divided into a convenient number of equal 
parts called degrees. If into one hundred degrees, it is found 
that each degree represents a volume equal to the y^Voir 
(.00367) part of the total volume occupied by the air at freez- 
iug-point ; i.e., the increase of volume from the temperature of 
freezing-point to that of the boiling-point of water = 0.367 of the 
volume at freezing, the pressure being the same, and even having 
any value whatever (as well as one atmosphere), within ordi- 
nary limits, so long as it is the same both at freezing and boil- 

604 



THERMOMETERS. 605 

ing. It must be understood, however, that by temjjerature of 
boiling is always meant that of water boiling under one at- 
mosphere pressure. Another way of stating the above, if one 
hundred degrees are used between freezing and boiling, is as 
follows : That for each degree increase of temperature the in- 
crease of volume is -g^r °^ tlie tota l volume at freezing ; 273 
being the reciprocal of .00367. 

As it is not always practicable to preserve the pressure con- 
stant under all circumstances with an air- thermometer, we use 
the common mercurial thermometer for most practical pur- 
poses. In this, the tube is sealed at the outer extremity, with 
a vacuum above the column of mercury, and its indications 
agree very closely with those of the air-thermometer. That 
equal absolute increments of volume should imply equal incre- 
ments of heat imparted to these thermometric liuids (under 
constant pressure) could not reasonably be asserted without 
satisfactory experimental evidence. This, however, is not al- 
together wanting, so that we are enabled to say that within a 
moderate range of temperature equal increments of heat pro- 
duce equal increments of volume in a given mass not only of 
atmospheric air, but of the so-called " perfect" or " permanent" 
gases, oxygen, nitrogen, hydrogen, etc. (so named before it was 
found that they could be liquefied). This is nearly true for 
mercury also, and for alcohol, but not for water. Alcohol has 
never been frozen, and hence is used instead of mercury as a 
thermometric substance to measure temperatures below the 
freezing-point of the latter. 

The scale of a mercurial thermometer is fixed ; but with an 
air-thermometer we should have to use a new scale, and in a 
new position on the tube, for each value of the pressure. 

467. Thermometric Scales. —In the Fahrenheit scale the tube 
between freezing and boiling is marked off into 180 equal 
parts, and the zero placed at 32 of these parts below the freez- 
ing-point, which is hence -f- 32°, and the boiling-point -\- 212°» 

The Centigrade, or Celsius, scale, which is the one chiefly 
used in scientific practice, places its zero at freezing, and 100° 
at boiling-point. Hence to reduce 



606 MECHANICS OF ENGINEERING. 

Fahr. readings to Centigrade, subtract 32° and multiply by |; 
Cent. " " Fahrenheit, multiply by ■§- and add 32°. 

468. Absolute Temperature. — Experiment also shows that if 
a mass of air or other perfect gas is confined in a vessel whose 
volume is but slightly affected by changes of temperature, 
equal increments of temperature (and therefore equal incre- 
ments of heat imparted to the gas, according to the preceding 
paragraph) produce equal increments of tension (i.e., pressure 
per unit area) ; or, as to the amount of the increase, that when 
the temperature is raised by an amount 1° Centigrade, the ten- 
sion is increased -^k^ °^ ^ s vame a ^ freezing-point. Hence, 
theoretically, an ideal barometer (containing a liquid unaffected 
by changes of temperature) communicating with the confined 
gas (whose volume practically remains constant) would by 
its indications serve as a thermometer, 
Fig. 519, and the attached scale could be 
graduated accordingly. Thus, if the col- 
imn stood at A when the temperature 
was freezing, A would be marked 0° on 
^ the Centigrade system, and the degree 
^3 spaces above and below A would each 
FlG - 519 - = A- of the height AB, and therefore 

the point B (cistern level) to which the column would sink if 
the gas-tension were zero would be marked — 273° Centi- 
grade. 

But a zero-pressure, in the Kinetic Theory of Gases (§ 408), 
signifies that the gaseous molecules, no longer impinging 
against the vessel walls (so that the press. = 0), have become 
motionless; and this, in the Mechanical Theory of Heat, or 
Thermodynamics, implies that the gas is totally destitute of heat. ' 
Hence this ideal temperature of — 273° Centigrade, or — 460° 
Fahrenheit, is called the Absolute Zero of Temperature, and by 
reckoning temperatures from it as a starting-point, our formulae 
will be rendered much more simple and compact. Tempera- 
ture so reckoned is called absolute temperature, and will be 
denoted by the letter T. Hence the following rules for re- 
duction : 



GASES AND VAPORS. 607 

Absol. temp. Tin Cent, degrees == Ordinary Cent, -f- 273° ; 
Absol. temp. T in Fahr. degrees = Ordinary Falir. -f- 460°. 
Tor example, for 20° Cent., T = 293° Abs. Cent. 

469. Distinction Between Gases and Vapors. — All known 
gases can be converted into liquids by a sufficient reduction of 
temperature or increase of pressure, or both ; some, however, 
with great difficulty, such as atmospheric air, oxygen, hydro- 
gen, nitrogen, etc., these having been but recently (1878) re- 
duced to the liquid form. A vapor is a gas near the point of 
liquefaction, and does not show that regularity of behavior 
under changes of temperature and pressure characteristic of a 
gas when at a temperature much above the point of liquefac- 
tion. All gases treated in this chapter (except steam) are sup- 
posed in a condition far removed from this stage. The fol- 
lowing will illustrate the properties of vapors. See Fig. 520. 
Let a quantity of liquid, say water, be intro- THERM . 
duced into a closed space, previously vacuous, 
of considerably larger volume than the water, 
and furnished with a manometer and ther- 
mometer. Vapor of water immediately be- 
gins to form in the space above the liquid, and 
continues to do so until its pressure attains a , JnhiAu/}'/'A 
definite value dependent on the temperature, FlG ' 520 ' 
and not on the ratio of the volume of the vessel and the origi- 
nal volume of water ; e.g., if the temperature is 70° Fahren- 
heit, the vapor ceases to form when the tension reaches a value 
of 0.36 lbs. per sq. inch. If heat be gradually applied to raise 
the temperature, more vapor will form (with ebullition ; i.e., 
from the body of the liquid, unless the heat is applied very 
slowly), but the tension will not rise above a fixed value for 
each temperature (independent of size of vessel) so long as 
there is any liquid left. Some of these corresponding values, 
for water, are as follows : For a 

Fahr. temp. = 70° 100° 150° 212° 220° 287° 300° 

Te ^Tc!! ( !^l = - 36 °- 93 3.69 14.7 17.2 55.0 67.2 
per sq. m. \ \ 

= one atm. 
At any such stage the vapor is said to be saturated. 




MECHANICS OF ENGINEEKING. 



Finally, at some temperature, dependent on the ratio of the 
original volume of water to that of the vessel, all of the water 
will have been converted into vapor (i.e., steam); and if the 
temperature be still further increased, the tension also increases 
and no longer depends on the temperature alone, hut also on 
the heaviness of the vapor when the water disappeared. The 
vapor is now said to be superheated, and conforms more in its 
properties to perfect gases. 



470. Critical Temperature. — From certain experiments there 
seems to be reason to believe that at a certain temperature, 
called the critical temperature, different for different liquids, 
all of the liquid in the vessel (if any remains, and supposing 
the vessel strong enough to resist the pressure) is converted 
into vapor, whatever be the size of the vessel. That is, above 
the critical temperature the substance is necessarily gaseous, 
in the most exclusive sense, incapable of liquefaction by pres- 
sure alone ; while below this temperature it is a vapor, and lique- 
faction will begin if, by compression in a cylinder and conse- 
quent increase of pressure, the tension can be raised to a value 
corresponding, for a state of saturation, to the temperature 
(in such a table as that just given for water). For example, if 
vapor of water at 220° Fahrenheit and tension of 10 lbs. per 
sq. inch (this is superheated steam, since 220° is higher than 
the temperature which for saturation corresponds to p = 10 
lbs. per sq. inch) is compressed slowly (slowly, to avoid change 
of temperature) till the tension rises to 17.2 lbs. per sq. in., 
which (see above table) is the pressure of saturation for a tem- 
perature of 220° Fahrenheit for water-vapor, the vapor is satu- 
rated, i.e., liquefaction is ready to begin, and during any fur- 
ther slow reduction of volume the pressure remains constant 
and some of the vapor is liquefied. 

By " perfect gases," or gases proper, we may understand, 
therefore, those which cannot be liquefied by pressure unac- 
companied by great reduction of temperature; i.e., whose 
" critical temperatures" are very low. The critical temperature 
of 1ST 2 0, or nitrous oxide gas, is between — 11° and -|- 8° Cen- 
tigrade, while that of oxygen is said to be at — 118° Centi- 



LAW OF CHAELES. 609 

grade. [See p. 471, vol. 122 of the Journal of the franklin 
Institute. For an account of the liquefaction of oxygen, etc., 
see the same periodical, January to June, 1878.] 

471. Law of Charles (and of Gay Lussac). — The mode of gradu- 
ation of the air-thermometer may be expressed in the follow- 
ing formula, which holds good (for practical purposes) within 
the ordinary limits of experiment for a given mass of any 

perfect gas, the tension remaining constant : 

V = V + 0.00367 V t = V (l + .003670 ;••(*) 

in which V denotes the volume occupied by the given mass 
at freezing-point under the given pressure, V its volume at 
any other temperature t Centigrade under the same pressure, 
Now, 273 being the reciprocal of .00367, we may write 

(273+Q . V_ T_ (press.) . m 

V ~ K 273 ' 1 ' 6 '' V~T a • • J const, p ^ 

(see § 468 ;) in which T = the absolute temperature of freezing- 
point, = 273° absolute Centigrade, and T the absolute tem- 
perature corresponding to t Centigrade. Eq. (2) is also true 
when T and T are both expressed in Fahrenheit degrees (from 
absolute zero, of course). Accordingly, we may say that, the 
pressure remaining the same, the volume of a given mass of 
gas varies directly as the absolute temperature. - 

Since the weight of the given mass of gas is invariable at a 
given place on the earth's surface, we may 

always use the equation Vy = Y y , (3) 

pressure constant or not, and hence (2) may be rewritten 

y T 

— = -^7-. . . (press, const.) ; . (4) 

i.e., if the pressure is constant, the heaviness {and therefore 
the specific gravity) varies inversely as the absolute tempera- 
ture. 



MECHANICS OF ENGINEEKING. 



Experiment also shows (§ 468) that if the volume [and there- 
fore the heaviness, eq. (3)] remains constant, while the tem- 
perature varies, the tension p will change according to the 
following relation, in which p = the tension when the tem- 
perature is freezing: 



P=Po + 



i , 273 + t 



(5) 



t denoting the Centigrade temperature 
as before, we have 

J - L 
p. t; • 



Hence transforming, 



vol., and .*. 
heav., const. r 



(6) 



or, the volume and heaviness remaining constant, the tension 
of a given mass of gas varies directly as the absolute tempera- 
ture. This is called the Law of Charles (or of Gay Lussac). 



472. General Formulae for any Change of State of a Perfect Gas. 

— If any two of the three quantities, viz., volume (or heavi- 
ness), tension, and temperature, are changed, the new value of 
the third is determinate from those of the other two, according 
B to a relation proved as follows (remember- 
ing that henceforth the absolute temperature 
only will be used, T, § 468) : Fig. 521. 
At A a certain mass of gas at a tension of 
p Q , one atmosphere, and absolute tempera- 
ture T (freezing), occupies a volume V Q . 
Let it now be heated to an absolute temp. 
= T' ', without change of tension (expanding 
behind a piston, for instance). Its volume will increase to a 
value V which from (2) of § 471 will satisfy the relation 



v , 

3,' T 0' 




V, 








c 


V, 
p,T. 





Fig. 521. 



V 

v. 



2^ 

t; 



CD 



(See B in figure.) 

Let it now be heated without change of volume to an abso- 
lute temperature T (C in figure). Its volume is still V, but 



LAWS OF PERFECT GASES. 611 

the tension has risen to a value p, such that, on comparing B 
and C by eq. (6), we have 

£=f « 

Combining (7) and (8), we obtain for any state in which the 
tension is^>, volume V, and absolute temperature T, in 

7) V 7) 1 \7~ 7) V^ 

{General). . . 1 -—-— 1 -^—^\ or *-=- = a constant ; . (9) 

-Z .x J. 

or 

(GWaZ). • • • & ~ f = £ ^, (10) 

which, since 

{General). . Vy = V y 9 = V m y m = V n y n , . . . (11) 

is true for any change of state, we may also write 

(General) .... JL = _^_, (12) 

or 

yir ~ VT ( } 

These equations (9) to (13), inclusive, hold good for any state 
of a mass of any perfect gas (most accurately for air). The 
subscript refers to the state of one-atmosphere tension and 
freezing-point temperature, m and n to any two states what- 
ever (within practical limits) ; y is the heaviness, §§ 7 and 409, 
and T the absolute temperature, § 468. 

lip, V, and T of equation (9) be treated as variables, and 
laid off to scale as co-ordinates parallel to three axes in space, 
respectively, the surface so formed of which (9) is the equation 
is a hyperbolic paraboloid. 

473. Examples. — Example 1. — What cubic space will be 
occupied by 2 lbs. of hydrogen gas at a tension of two atmos- 
pheres and a temperature of 27° Centigrade ? 



612 MECHANICS OF ENGINEEEING. 

With the inch-lb. -sec. system we have p = 14.7 lbs. per sq. 
inch, y = [.0056 -f- 1728] lbs. per cubic inch, and T = 273° 
absolute Centigrade, when the gas is at freezing-point at one 
atmosphere (i.e., in state sub-zero). In the state mentioned in 
the problem, we have^> = 2 X 14.7 lbs. per sq. in., 

T = 273 + 27 = 300° absolute Centigrade, 

while y is required. Hence, from eq. (12), 

2 X 14.7 14 1 7 . 

Y 300 ~~ (.0056 -v- 1728)273 ' 

,\ y = * ^ lbs. per cub. in. = .0102 lbs. per cub. foot ; and if 

1728 r r 

the total weight, = G, = Vy, is to be 2 lbs., we have (ft., lb., 
sec.) V= Z'-t- 0102 = 196 cubic feet.— Ans. 

Example 2. — A mass of air originally at 24° Centigrade 
and a tension indicated by a barometric column of 40 inches 
of mercury has been simultaneously reduced to half its 
former volume and heated to 100° Centigrade ; required its 
tension in this new state, which we call the state n, m being the 
original state. Use the inch, lb., sec. We have given, there- 
fore, p m = |f X 14.7 lbs. per sq. inch, T m = 273 + 24 = 297° 
absolute Centigrade, the ratio 

V m : V n = 2 : 1, and T n = 273 + 100 =373° Abs. Cent.; 
whiles is the unknown quantity. From eq. (10), hence, 

^ = -^.^-.^=2xHf.«Xl4.7 = 49.221bs.persq.in., 

which an ordinary steam-gauge would indicate as 
(49.22 — 14.7) = 34.52 lbs. per sq. inch. 

(That is, if the weather barometer indicated exactly 14.7 lba 
per sq. inch.) 



EXAMPLES. PERFECT GASES. 



613 



Example 3. — A mass of air, Fig. 522, occupies a rigid closed 
vessel at a temperature of 15° Centigrade (equal to that of sur- 



& 



££?M*l*{\W 




Fig. 522. 



rounding objects) and a tension 
of four atmospheres [state m]. 
By opening a stop-cock a few 
seconds, thus allowing a portion 
of the gas to escape quickly, and 
then shutting it, the remainder 
of the air [now in state ri] is found to have a tension of only 
2.5 atmospheres (measured immediately) ; its temperature can- 
not be measured immediately (so much time being necessary 
to affect a thermometer), and is less than before. To compute 
this temperature, T n , we allow the air now in the vessel to 
come again to the same temperature as surrounding objects 
(15° Centigrade) ; find then the tension to be 2.92 atmospheres. 
Call the last state, state r (inch, lb., sec). The problem then 
stands thus : 



p m =4x14.7 

T m = 288° Abs. Cent. 



p n = 2.5 X 14.7 

jr _ o j principal 
n ~ ) unknown 



p r = 2.92 X 14.7 

yr = Yn (since Vr = Vn) 

T r = T m = 288° Abs. Cent. 



In states n and r the heaviness is the same ; hence an equa- 
tion like (6) of § 471 is applicable, whence 

frW orr„ = g^x288=WAb,Ce,t. 



or — 27° Centigrade ; considerably below freezing, as a result of 
allowing the sudden escape of a portion of the air, and the con- 
sequent sudden expansion, and reduction of tension, of the re- 
mainder. In this sudden passage from state m to state n, the 
remainder altered its heaviness (and its volume in inverse ratio) 
in the ratio (see eqs. (11) and (10) of § 472) 



't _ 


_ ' m J^n J- m 


2.5 X 14.7 


'm 


* n j/m, •*■ n 


4 X 14.7 



288 
246 



= 0.73. 



Now the heaviness in state m (see eq. (12), § 472) was 



614 



MECHANICS OF ENGINEERING. 



Vm = 



_Pm y Q T _±xU.7 .0807 273 



.306 



i>0 



288 



1728 14.7 1728 



per cub. in. = .306 lbs. per cub. ft. 



0.73 Xy m = 0.223 lbs. per cub. ft., 

of the original quan- 



lbs, 



and also, since V m = 0.73 V n , about T \ 
tity of air in vessel has escaped. 

[Note. — By numerous experiments like this, the law of 
cooling, when a mass of gas is allowed to expand suddenly (as, 
e.g., behind a piston, doing work) has been determined ; and 
vice versa, the law of heating under sudden compression 
§ 487.] 



see 





*•*>. ■ 




i 




^ = 




—"*-*- 





474. The Closed Air-manometer. — If a manometer be formed 
of a straight tube of glass, of uniform cylindrical bore, which 
is partially filled with mercury and then inverted in a cistern 
of mercury, a quantity of air having been left between the 

mercury and the upper end of the 
tube, which is closed, the tension of 
this confined air (to be computed 
from its observed volume and tem- 
perature) must be added to that due 
to the mercury column, in order to 
obtain the tension p' to be measured. 
See Fig. 523. The advantage of this 
kind of instrument is, that to meas- 
ure great tensions the tube need not 
be very long. Let the temperature 
T y of whole instrument, and the tension p x of the air or gas 
in the cistern, be known when the mercury in the tube stands 
at the same level as that in the cistern. The tension of the 
air in the tube must now be^ also, its temperature T x , and its 
volume is V 1 = Fh x , F being the sectional area of the bore of 
the tube ; see on left of figure. "When the instrument is used, 
gas of unknown tension p' is admitted to the cistern, the tem- 
perature of the whole instrument being noted (= T\ and the 
heights h and h" are observed (h -\- h" cannot be put = A, 



p^\ i 



Fig. 523. 



CLOSED AIR-MANOMETER. 615 

unless the cistern is very large), p' is then computed as fol- 
lows (eq. (2), § 413) : 

p'=h"y m +p; (1) 

in which p = the tension of the air in the tube, and y m the 
heaviness of mercury. But from eq. (10), § 472, putting 
V l = Fh 1 and V=Fh, 

Hence finally, from (1) and (2), 

j^V'rm+.^.jrP* (3) 

Since T l9 p ti and h r are fixed constants for each instrument, 
we may, from (3), compute p / for any observed values of h and 
^(N.B. T and T t are absolute temperatures), and construct 
a series of tables each of which shall give values of p' for a 
range of values of A, and one special value of T. 

Example. — Supposing the fixed constants of a closed air* 
manometer to be (in inch-lb.-sec. system) p t — 14.7 (or one 
atmosphere), T x = 285° Abs. Cent, (i.e., 12° Centigrade), and 
A, = 3' 4" = 40 inches ; required the tension in the cistern 
indicated by h" = 25 inches and h = 15 inches, when the 
temperature is — 3° Centigrade, or J= 270° Abs. Cent. 

For mercury, y m = [848.7 -r- 1728] (§ 409) (though strictly 
it should be specially computed for the temperature, since it 
varies about .00002 of itself for each Centigrade degree). 
Hence, eq. (3), 

lbs. per sq. inch, or nearly 3^ atmospheres [steam-gauge would 
read 34.7 lbs. per sq. in.]. 

475. Mariotte's Law, (or Boyle's,) Temperature Constant ; i.e., 
Isothermal Change. — If a mass of gas be compressed, or al- 



616 



MECHANICS OF ENGINEERING. 



lowed to expand, isothermally, i.e., without change of tem- 
perature (practically this cannot be done unless the walls of the 
vessel are conductors of heat, and then the motion must be 
slow), eq. (10) of § 472 now becomes (since T m = T n ) 



j Mariotte's Law, 
\ Temp, constant 



* mjPm — * nJP% 



or^ = £;.(l) 



i.e., the temperature remaining unchanged, the tensions are 
inversely proportional to the volumes, of a given mass of a 
perfect gas / or, the product of volume by tension is a constant 
quantity. Again, since V m y m = V n y n for any change of 
state, 



Mariotte's Law, 
Temp, constant 



Pm__Ym 



or 



Pr 

Y% 



*T\ ' (3) 



i.e., the pressures (or tensions are directly proportional to the 
(first power of the) heavinesses, if the temperature is the same. 
This law, which is very closely followed by all the perfect 
gases, was discovered by Boyle in England and Mariotte in 
France more than two hundred years ago, but of course is only 
a particular case of the general formula, for any change of 
state, in § 472. It may be verified experimen- 
tally in several ways. E.g., in Fig. 524, the 
tube M being closed at the top, while PN is 
open, let mercury be poured in at P until it 
reaches the level A' B' . The air in OA' is now 
at a tension of one atmosphere. Let more mer- 
cury be slowly poured in at P, until the air 
confined in has been compressed to a volume 
OA" = i of OA f , and the height B"E" then 
measured ; it will be found to be 30 inches ; i.e., 
the tension of the air in is now two atmos- 
pheres (corresponding to 60 inches of mercury). 
Again, compress the air in to \ its original 
volume (when at one atmosphere), i.e., to volume OA'" = 
\OA \ and the mercury height B'"E'" will be 60 inches, show- 
ing a tension of three atmospheres in the confined air at (90 



II 
B 

b' 


- 


f 

e"' 

b'" 
A 


o 


-^r 


— 




=j^—= 


N 


■=-.= ■=- 



Fig. 524. 



617 

inches of mercury in a barometer). It is -understood that the 
temperature is the same, i.e., that time is given the compressed 
air to acquire the temperature of surrounding objects after 
being heated by the compression, if sudden. 

[JSTote. — The law of decrease of steam -pressure in a steam- 
engine cylinder, after the piston has passed the point of " cut- 
off " and the confined steam is expanding, does not materially 
differ from Mariotte's law, which is often applied to the case 
of expanding steam ; see § 479.] 

While Mariotte's law may be considered exact for practical 
purposes, it is only approximately true, the amount of the 
deviations being different at different temperatures. Thus, 
for decreasing temperatures the product Vp of volume by 
tension becomes smaller, with most gases. 

Example 1. — If a mass of compressed air expands in a 
cylinder behind a piston, having a tension of 60 lbs. per sq. 
inch (45.3 by steam-gauge) at the beginning of the expansion, 
which is supposed slow (that the temperature may not fall) ; 
then when it has doubled in volume its tension will be only 
30 lbs. per sq. inch ; when it has tripled in volume its tension 
will be only 20 lbs. per sq. inch, and so on. 

Example 2. Diving-hell. — Fig. 525. If the cylindrical 
diving-bell AB is 10 ft. in height, in what 
depth, h = ?, of salt water, can it be let down ■;.-."■.": 
to the bottom, without allowing the water to i\^- 
rise in the bell more than a distance a = 4 ft.? l':^ 
Call the horizontal sectional area, F. The ^^_- 
mass of air in the bell is constant, at a constant -%^- 
temperatnre. First, algebraically ; at the 
surface this mass of air occupied a volume 
V m = Fh" at a tension p m = 14.7 X 144 lbs. 
per sq. ft., while at the depth mentioned it is 
compressed to a volume V n = Fi]i" — a\ and 



a - 



is at a tension p n =p m + (h — a)y w , in which &/y/////////////J 
y w — heaviness of salt water. Hence, from **<*- 525 - 

eq. (1), 

p m Fh"=\p m + {Ti~a) Ya ]F{h"-a); . . (3) 



618 MECHANICS OF ENGINEERING. 



•• A = a [ 1 + (F^] ; « 



hence, numerically, (ft., lb., sec.,) 



h = 4 X [~1 



14.7 X 144 i 
(10 — 4) X 64 



= 26.05 feet. 



476. Mixture of Gases. — It is sometimes stated that if a vessel 
is occupied by a mixture of gases (between which there is no 
chemical action), the tension of the mixture is equal to the sum 
of the pressures of each of the component gases present; or, 
more definitely, is equal to the sum of the pressures which the 
separate masses of gas would exert on the vessel if each in turn 
occupied it alone at the same temperature. 

This is a direct consequence of Mariotte's law, and may be 
demonstrated as follows : 

Let the actual tension be^>, and the capacity of the vessel V. 
Also let V l9 F 2 , etc., be the volumes actually occupied by the 
separate masses of gas, so that 

r, + F,+ ...= V; (1) 

and p l9 p 3 , etc., the pressures they would individually exert 
when occupying the volume V alone at the same tempera- 
ture. Then, by Mariotte's law, 

Vp^V lP ; Vj>,= Vj>; etc.; . . . (2) 

whence, by addition, we have 

i.e., p=± Pl + p% + . (3) 

Of course, the same statement applies to any number of 
separate parts into which we may imagine a mass of homo- 
geneous gas to be divided. 

For numerical examples and practical questions in the solu- 
tion of which this principle is useful, see p. 239, etc., Ean- 
kine's Steam-engine. (Rankine uses 0.365, where 0.367 has 
been used here.) 






BAROMETRIC LEVELLING. 619 

477. Barometric Levelling. — By measuring with a barometer 
the tension of the atmosphere at two different levels, simul- 
taneously, and on a still day, the two localities not being widely 
separated horizontally, we may compute their vertical distance 
apart if the temperature of the stratum of air between them 
is known, being the same, or nearly so, at both m 

stations. Since the heaviness of the air is ^j \ ;'•;'! - ; .. f" 
different in different layers of the vertical { '.' t k \ ".'.-' \ ••'*. j '. 
column between the two elevations i^and M, '.'.' \ \ 'i\ \. . • . \ \' 
Fig. 526, we cannot immediately regard the ' tfjff7|T*Ap 
whole of such a column as a free body (as was ,. • L!7y;|~~-P + f^ 
done with a liquid, § 412), but must consider 
a horizontal thin lamina, Z, of thickness 
— dz and at a distance = z (variable) below 
M, the level of the upper station, N being 
the lower level at a distance, A, from M. 

The tension, p, must increase from M 
downwards, since the lower laminae have to support a greater 
weight than the upper ; and the heaviness y must also increase, 
proportionally toj9, since we assume that all parts of the col- 
umn are at the same temperature, thus being able to apply 
Mariotte's law. Let the tension and heaviness of the air at 
the upper base of the elementary lamina, Z, be p and y re- 
spectively. At the lower base, a distance dz below the upper, 
the tension \s>p -\- dp* I^t the area of the base of lamina be 
F\ then the vertical forces acting on the lamina are Fp, down- 
ward ; its weight yFdz downward ; and F(p + dp) upward. 
For its equilibrium ^(vert. compons.) must = 0; 

.-. F(p + dp)-Fp- Fydz = ; 

i.e., dp = ydz, (1) 

which contains three variables. But from Mariotte's law, 
§ 475, eq. (2), if p n and y n refer to the air at iV, we may 

substitute y = — p and obtain, after dividing by p, to separate 

the variables^? and s, 



620 MECHANICS OF ENGINEERING. 

*!.#'=» . . (2) 

Yn JP 

Summing equations like (2), one for each lamina between 
M (where p =p m and z = 0) and iV (where p =^> n ands == A), 
we have 



Le.,A = ^L log.. 

Yn 






(3) 



which gives A, the difference of level, or altitude, between M 
and iV 7 , in terms of the observed tensions p n andj? TO , and of y n , 
the heaviness of the air at JV, which may be computed from 
eq. (12), § 472, substituting from which we have finally 

A=f.^.log..[f:T, .... (4) 

/ o -*■ o V-J/m-1 

in which the subscript refers to freezing-point and one at- 
mosphere tension ; T n and T Q are absolute temperatures. For 
the ratio p n : p m we may put the equal ratio h n : h m of the 
actual barometric heights which measure the tensions. The 
log. .(or Naperian, or natural, or hyperbolic, log.) = (common 
log. to base 10) X 2.30258, From § 394, y of air = 0.08076 
lbs. per cub. ft., and p = 14.701 lbs. per sq. inch ; T == 273° 
Abs. Cent. 

If the temperature of the two stations (both in the shade) 
are not equal, a mean temp. = i(T m -\- T n ) may be used for 
T n in eq. (4), for approximate results. Eq. (4) may then be 
written 

A (in feet) = 26213 ^ . log., f^l .... (5) 

The quantity ^ = 26213 ft., just substituted, is called the 

height of the homogeneous atmosphere, i.e., the ideal height 
which the atmosphere would have, if incompressible and non- 



BAROMETKIC LEVELLING — ADIABATIC LAW. 621 

expansive like a liquid, in order to exert a pressure of 14.701 
lbs. per sq. inch upon its base, being throughout of a constant 
heaviness = .08076 lbs. per cub. foot. 
By inversion of eq. (4) we may also write 

p m e» Tn =p n , (6) 

where e = 2.71828 = the Naperian Base, which is to be raised 

v T 

to the power whose index is the abstract number — . -=f . A, 

and the result multiplied by p m to obtain p n . 

Example. — Having observed as follows (simultaneously) : 

At lower station iV, h n = 30.05 in. mercury ; temp. = 77.6° F. ; 
« upper " M, h m = 23.66 " " ' " = 70.4° F. ; 

required the altitude h. From these figures we have a mean 
absolute temperature of 460° + J(77.6 + 70.4) = 534° Abs. 
Fahr. ; hence, from (5), 



h = 26213 X 44* X 



2.30258 X log. 10 [|g|] = 6787.9 ft. 



(Mt. Guanaxuato, in Mexico, by Baron von Humboldt.) 
Strictly, we should take into account the latitude of the place, 
since y varies with g (see § 76), and also the decrease in the 
intensity of gravitation as we proceed farther from the earth's 
centre, for the mercury in the barometer weighs less per cubic 
inch at the upper station than at the lower. 

Tables for use in barometric levelling can be found in Traut- 
wine's Pocket-book, and in Searles's Field-book for Railroad 
Engineers, as also tables of boiling-points of water under dif- 
ferent atmospheric pressures, forming the basis of another 
method of determining heights. 

478. Adiabatic Change — Poisson's Law. — By an adiabatic 
change of state, on the part of a gas, is meant a compression 
or expansion in which work is done vpon the gas (in compress- 



622 MECHANICS OF ENGINEERING. 

ing it) or by the gas (in expanding against a resistance) when 
there is no transmission of heat between the gas and enclosing 
vessel, or surrounding objects, by conduction or radiation. 
This occurs when the volume changes in a vessel of non-cob 
ducting material, or when the compression or expansion takes 
place so quickly that there is no time for transmission of heat 
to or from the gas. 

The experimental facts are, that if a mass of gas in a cylinder 
be suddenly compressed to a smaller volume its temperature is 
raised, and its tension increased more than the change of vol- 
ume would call for by Mariotte's law ; and vice versd, if a gas 
at high tension is allowed to expand in a cylinder and drive a 
piston against a resistance, its temperature falls, and its tension 
diminishes more rapidly than by Mariotte's law. 

Again (see Example 3, § 473), if ^\ of the gas in a rigid 
vessel, originally at 4 atmos. tension and temperature of 
15° Cent., is allowed to escape suddenly through a stop-cock 
into the outer air, the remainder, while increasing its volume 
in the ratio 100 : 73, is found to have cooled to — 27° Cent., 
and its tension to have fallen to 2.5 atmospheres; whereas, by 
Mariotte's law, if the temperature had been kept at 288° Abs. 
Cent., the tension would have been lowered to ^ 3 7 of 4, i.e., 
to 2.92 atmospheres only. 

The reason for this cooling during sudden expansion is, ac- 
cording to the Kinetic Theory of Gases, that since the " sensi- 
ble heat" (i.e., that perceived by the thermometer), or u hot- 
ness" of a gas depends on the velocity of its incessantly moving 
molecules, and that each molecule after impact with a receding 
piston has a less velocity than before, the temperature neces- 
sarily falls; and vice versa, when an advancing piston com- 
presses the gas into a smaller volume. 

If, however, a mass of gas expands without doing work, as 
when, in a vessel of two chambers, one a vacuum, the other 
full of gas, communication is opened between them, and the 
gas allowed to fill both chambers, no cooling is noted in the 
mass as a whole (though parts may have been cooled tem- 
porarily). 

By experiments similar to that in Example 3, § 473, it has 



ADIABATIC CHANGE — EXAMPLES. 623 

been found that for air and the " perfect gases," in an adiabatic 
change of volume [and therefore of heaviness], the tension 
varies inversely with the 1.41th power of the volume. This 
is called Poissorfs Law. For ordinary purposes (as Weisbach 
suggests) we may use f instead of 1.41, and hence write 

Adiabat. \ Pm _ ty^f nr Pm-(Vn \* m 

Change] ' ' T^KyJ ' * p n ~WJ> ' ' W 

and combining this relation with the general eqs. (10) and (13), 
§ 472, we have also 

Adiabat. \ Pm _ / TmY / 9 \ 

Change \ p.~\TJ' K > 

i.e., the tension varies directly as the cube of the absolute tem- 
perature; also, 



Adiabat. \ l^A - ( T »\* nr Y n - fZ*-\". 
Change] ' \ VJ ~ \Tj ' y m ~ \T m )" ' 



. (3) 



i.e., the volume is inversely, and the heaviness directly, as the 
square of the absolute temperature. 

Here m and n refer to any two adiabatically related states. 
Tis the absolute temperature. 

Example 1. — Air in a cylinder at 20° Cent, is suddenly 
compressed to -J- its original volume (and therefore is six times 
as dense, i.e., has six times the heaviness, as before). To what 
temperature is it heated ? Let m be the initial state, and n the 
final. From eq. (3) we have 

or nearly double the absolute temperature of boiling water. 

Example 2. — After the air in Example 1 has been given 
time to cool again to 20° Cent, (temperature of surrounding 
obiects") it is allowed to resume, suddenly, its first volume, i.e., 



624 MECHANICS OF ENGINEERING. 

to increase its volume sixfold by expanding behind a piston. 
To what temperature has it cooled ? Here T m = 293° Abs„ 
Cent., the ratio V m : V n == -J-, and T n is required. Hence, 
from (3), 

*| = JT ; ... T n = 293 - VT= 119.5° Abs. Cent., 

or = — 154° Cent., indicating extreme cold. 

From these two examples the principle of one kind of ice- 
making apparatus is very evident. As to the work necessary 
to compress the air in Example 1, see § 483. It is also evident 
why motors using compressed air expansively have to encoun- 
ter the difficulty of frozen watery vapor (present in the air to 
some extent). 

Example 3. — What is the tension of the air in Example 1 
(suddenly compressed to -§- its original volume) immediately 
after the compression, if the original tension was one atmos- 
phere ? That is, with V n : V m = 1 : 6, and^> TO = 14.7 lbs. per 
sq. inch, p n = % From eq. (1), (in., lb., sec.,) 

p n = 14.7 X 6 f = 14.7 V216 = 216 

lbs. per sq. inch ; whereas, if, after compression and without 
change of volume, it cools again to 20° Cent., the tension is 
only 14.7 X 6 = 88.2 lbs. per sq. inch (now using Mariotte's 
law). 

479. Work of Expanding Steam following Mariotte's Law. — 
Although gases do not in general follow Mariotte's law in ex- 
panding behind a piston (without special provision for sup 
plying heat), it is found that the tension of saturated steam 
(i.e., saturated at the beginning of the expansion) in a steals 
engine cylinder, when left to expand after the piston has 
passed the point of " cut-off" diminishes very nearly in 
accordance with Mariotte's law, which may therefore be ap- 
plied in this case to find the work done per stroke, and thence 
tie power. In Fig. 527 a horizontal steam-cylinder is 



EXPANDING STEAM. 



625 




shown in which the piston is making its left-to-right stroke. 
The "back-pressure" is con- 
stant and = Fq, F being the 
area of the piston and q the 
intensity (i.e., per unit area) 
of the back or exhaust pres- 
sure on the right side of the 
piston ; while the forward 
pressure on the left face of the 
piston = Fp, in which p is the 
steam-pressure per unit area, 
and is different at different 
points of the stroke. While the 
piston is passing from O" to 
D",p is constant, being —p b = the boiler-pressure, since the 
steam-port is still open. Between D" and C" , however, the 
steam being cut off (i.e., the steam-port is closed) at D' ', a dis- 
tance a from 0'\ p decreases with Mariotte's law (nearly), and 
its value is (Fa -i- Fx)p b at any point on CD", x being the 
distance of the point from 0" . 

Above the cylinder, conceive to be drawn a diagram in 
which an axis <9Xis || to the cylinder-axis, OY an axis 1 to 
the same, while is vertically above the left-hand end of the 
cylinder. As the piston moves, let the value of p correspond- 
ing to each of its positions be laid off, to scale, in the vertical 
immediately above the piston, as an ordinate from the axis X. 
Make OD' = q by the same scale, and draw the horizontal 
D ' C. Then the effective work done on the piston-rod while 
it moves through any small distance dx is 

d W = force X distance = F(p — q)dx, 



and is proportional to the area of the strip ES, whose width is 
dx and length =p — q ; so that the effective work of one 
stroke is 



c 



C" px=l 

W= dW 

vt V n 



/X=l 
(P- 
n 



q)dw, 



(1) 



626 MECHANICS OF ENGINEEK1NG. 

and is represented graphically by the area A'ARBC'D' A' 
Erom 0" to D" p is constant and =p b (while q is constant at 
all points), and x varies from to a ; 

••• \_f=F(p>-i)f ax = ili»-q>, ■ ■ (2) 

which may be called the work of entrance , and is represented 
by the area of the rectangle A' ADD'. 

From D" to C"p is variable and, by Mariotte's law, = -p b ; 

X 

i.e., 

U'=Jf[«p k log..(i)-^-a)] ... (3) 

= the work of expansion, adding which to that of entrance, 
we have for the total effective work of one stroke 

W=Fp b a\ L l + log. e (^Jj-Fql ... (4) 

By effective work we mean that done upon the piston-rod 
and thus transmitted to outside machinery. Suppose the 
engine to be " double-acting" ; then at the end of the stroke a 
communication is made, by motion of the proper valves, be- 
tween the space on the left of the piston and the condenser of 
the engine ; and also between the right of the piston and the 
boiler (that to the condenser now being closed). On the return 
stroke, therefore, the conditions are the same as in the forward 
stroke, except that the two sides of the piston have changed 
places as regards the pressures acting on them, and thus the 
same amount of effective work is done as before. 

If n revolutions of the fly-wheel are made per unit of time 
(two strokes to each revolution), the effective work done per 
unit of time, i.e., the power of the engine, is 

L = 2n TF.-r-- 2/^^Jl + log.. Q] - qlj. (5) 






WOKK OF STEAM-ENGINES. 627 

For simplicity the above theory has omitted the considera- 
tion of " clearance" that is, the fact that at the point of " cut- 
off " the mass of steam which is to expand occupies not only 
the cylindrical volume Fa, but also the u clearance" or small 
space in the steam-passages between the valve and the entrance 
of the cylinder, the space between piston and valve which is 
never encroached upon by the piston. " Wire-drawing" has 
also been disregarded, i.e., the fact that during communication 
with the boiler the steam-pressure on the piston is a little less 
than boiler-pressure. For these the student should consult 
special works, and also for the consideration of water mixed 
with the steam, etc. Again, a strict analysis should take into 
account the difference in the areas which receive fluid-pressure 
on the two sides of the piston. 

Example 1. — A reciprocating steam-engine makes 120 revo- 
lutions per minute, the boiler-pressure is 40 lbs. by the gauge 
(i.e., p h = 40 4- 14.7 = 54.7 lbs. per sq. inch), the piston area 
is F= 120 sq. in., the length of stroke I = 16 in., the steam 
being "cut off" at J stroke (.*. a = 4 in., and I : a = 4.00), 
and the exhaust pressure corresponds to a " vacuum of 25 
inches" (by which is meant that the pressure of the exhaust 
steam will balance 5 inches of mercury), whence q = -^ of 
14.7 = 2.45 lbs. per sq. inch. Required the work per stroke, 
W, and the corresponding power Z. 

Since I : a = 4, we have log. 6 4 = 2.302 X .60206 = 1.386, 
and from eq. (4), (foot, lb., sec.,) 

W=zffi (54.7 X 144) . i . [2.386] - iff (2.45 X 144) . f 

= 5165.86 - 392.0 = 4773.868 ft. lbs. of work per stroke, 
and therefore the power at 2 rev. per sec. (eq. 5) is 

L = 2 X 2 X 4773.87 = 19095.5 ft. lbs. per second. 

Hence in horse-powers, which, in ft.,-lb.-sec. system, =Z-5-550, 

Power = 19095.5 -~ 550 = 34.7 H. P. 

Example 2. — Eequired the weight of steam consumed per 



628 MECHANICS OF ENGINEERING. 

second by the above engine with given data ; assuming with 
Weisbach that the heaviness of saturated steam at a definite 
pressure (and a corresponding temperature, % 469) is about f of 
that of air at the same pressure and temperature. 

The heaviness of air at 54.7 lbs. per sq. in. tension and 
temperature 287° Fahr. (see table, § 469) would be, from eq. 
(12) of § 472 (see also § 409), 

v = rX P = -0807X492 ^1 = 019S 
7 T ' p 460 + 287 " 14.7 

lbs. per cub. foot, \ of which is 0.1237 lbs. per cub. ft. Now 
the volume* of steam, of this heaviness, admitted from the 
boiler at each stroke is V—Fa = ^.^ = 0.2777 cub. ft. y 
and therefore the weight of steam used per second is 

4 X .2777 X 0.1237 = 0.1374 lbs. 

Hence, per hour, 0.1374 X 3600 = 494.6 lbs. of feed-water 
are needed for the boiler. 

If, with this same engine, the steam is used at full boiler 
pressure throughout the whole stroke, the power will be 
greater, viz. = 2nFl(p b — q) — 33440 ft. lbs. per sec, but 
the consumption* of steam will be four times as great ; and 
hence in economy of operation it will be only 0.44 as efficient 
(nearly). 

480. Graphic Representation of any Change of State of a Con- 
fined Mass of Gas. — The curve of expansion AB in Fig. 527 is 
an equilateral hyperbola, the axes JTand Y being its asymp- 
totes. If compressed air were used instead of steam its ex- 
pansion curve would also be an equilateral hyperbola if its 
temperature could be kept from falling during the expansion 
(by injecting hot-water spray, e.g.), and then, following 
Mariotte's law, we would have, as for steam, (§ 475,) p V= con- 
stant, i.e., pFx = constant, and therefore px = constant, which 
is the equation of a hyperbola, p being the ordinate and x the 
abscissa. This curve (dealing with a perfect gas) is also called 
an isothermal, the x and y co-ordinates of its points being pro- 

* We here neglect the practical fact that a portion of the fresh steam 
entering the cylinder is condensed prematurely, so that the actual con- 
sumption is somewhat greater than as here derived. 



GRAPHICS OF CHANGE OF STATE OF GAS. 



629 




portional to the volume and tension, respectively, of a mass of 
air (or perfect gas) whose temperature is maintained constant. 
Hence, in general, if a mass of gas be confined in a rigid 
cylinder of cross-sec- 
tion F (area), provided 
with an air-tight pis- 
ton, Fig. 528, its vol- 
ume, Fx, is propor- 
tional to the distance 
OD — x (of the piston 
from the closed end of 
the cylinder) taken as 
an abscissa, while its 
tension p at the same 
instant may be laid off 
as an ordinate from D. 
Thus a point A is fixed. Describe an equilateral hyperbola 
through A, asymptotic to X and Y, and mark it with the ob- 
served temperature (absolute) of the air at this instant. In a 
similar way the diagram can be filled up with a great number 
of equilateral hyperbolas, or isothermal curves, each for its 
own temperature. Any point whatever (i.e., above the critical 
temperature) in the plane angular space TOX will indicate by 
its co-ordinates a volume and a tension, while the correspond- 
ing absolute temperature T will be shown by the hyperbola 
passing through the point, since these three variables always 
satisfy the relation (§ 472) 



*"■ 


x > 












1 


■ ■ - - 



Fig. 528. 



^=const.;i.e.,^=^. 



(1)' 



Any change of state of the gas in the cylinder may now be 
represented by a line in the diagram connecting the two points 
corresponding to its initial and final states. Thus, a point 
moving along the line AB, a portion of the isothermal marked 
293° Abs. Cent, represents a motion of the piston from D to 
F, and a consequent increase of volume, accompanied by just 
sufficient absorption of heat by the gas (from other bodies) to 
maintain its temperature at that figure (viz., its temperature at 



630 



MECHANICS OF ENGINEERING. 



A). If the piston move from D to E, without transmission 
of heat, i.e., adiabatically (§ 478), the tension falls more 
rapidly, and a point moving along the line AB' represents the 
corresponding continuous change of state. AB' is a portion 
of an adiabatic curve, whose equation, from § 478, is 



JL = [ Fxk 
p K V.FxJ 



or px§ = p K x K % = const. ; . (1) 



in which p K and x K refer to the point K where this particular 
adiabatic curve cuts the isothermal of freezing-point. Evi- 
dently an adiabatic may be passed through any point of the 
diagram. The mass of gas in the cylinder may change its 
state from A to B' by an infinite number of routes, or lines on 
the diagram, the adiabatic route, however, being that most likely 
to occur for a rapid motion of the piston. For example, we 
may cool it without allowing the piston to move (and hence 
without altering its volume nor the abscissa x) until the pres- 
sure falls to a value p B > = DL = EB\ and this change is rep- 
resented by the vertical path from A to L ; and then allow it 
to expand, and push the piston from D to E (i.e., do external 
work), during which expansion heat is to be supplied at just 
such a rate as to keep the tension constant, = ps' — Pl, this 
latter change corresponding to the horizontal path LB' from 
L to B'. 

It is further noticeable that the work done by the expanding 
gas upon the near face of the piston (or done upon the gas when 
compressed) when the space dx is described by the piston, is 
= Fpdx, and therefore is proportional to the area pdx of the 
small vertical strip lying between the axis X and the line or 
route showing the change of state ; whence the total work done 
on the near piston-face, being — Efpdx, is represented by the 
area fpdx of the plane figure between the initial and final 
ordinates, the axis X and the particular route followed be- 
tween the initial and final states (N.B. We take no account 
here of the pressure on the other side of the piston, the latter 
depending on the style of engine). For example, the work 
done on the near face of the piston during adiabatic expansion 



WORK OF ADIABATIC EXPANSION. 



631 



from D to E is represented by the plane figure AB'EDA, 
and is measured by its area. 

The mathematical relations between the quantities of heat 
imparted or rejected by conduction and radiation, and trans- 
formed into work, in the various changes of which the con- 
fined gas is capable, belong to the subject of Thermodynamics, 
which cannot be entered upon here. 

It is now evident how the cycle of changes which a definite 
mass of air or gas experiences when used in a hot-air engine, 
compressed-air engine, or air-compressor, is rendered more in- 
telligible by the aid of such a diagram as Fig. 528 ; but it 
must be remembered that during the entrance into, or exit 
from, the cylinder, of the mass of gas used in one stroke, the 
distance x does not represent its volume, and hence the locus 
of the points in the diagram determined by the co-ordinates p 
and x during entrance and exit does not indicate changes of 
state in the way just explained for the mass when confined in 
the cylinder. However, the work done by or upon the gas 
during entrance and exit will still be represented by the plane 
figure included by that locus (usually a straight horizontal 
line, pressure constant) and the axis of X and the terminal 
ordinates. 

481. Adiabatic Expansion in an Engine using Compressed Air. 
— Fig. 529. Let the compressed air at a tension j? m and an 
absolute temperature T m be supplied 
from a reservoir (in which the loss is 
continually made good by an air-com- y 
pressor). Neglecting the resistance of 
the port, its tension and temperature 
when behind the piston are still p m and 
T m . Let x n = length of stroke, and o 
let the cut-off (or closing of communi- | 
nation with the reservoir) be made at 
some point D where x = x m , the posi- 
tion of D being so chosen (i.e., the ratio 
x m : x n so computed) that after adia- 
batic expansion from D to E the pres- 
sure shall have fallen from p m at M {state m) to a value p n =p a 




632 



MECHANICS OF ENGINEERING. 



= one atmosphere at N(state n\ at the end of stroke ; so that 
when the piston returns the air will be expelled (" exhausted'') 
at a tension equal to that of the external atmosphere (though 
at a low temperature). Hence the back-pressure at all points 
either way will be = jp n per unit area of piston, and hence the 
total back-pressure = Fp n , F being the piston area. 

From to D the forward pressure is constant and = Fp m , 
and the effective work, therefore, or work on piston-rod from 
to D, is 



Work of entrance = W = F [p m — p n ~]x m , 
\-o 



a) 



represented by the rectangle M'MLN'. The cut-off being 
made at j9, the volume of gas now in the cylinder, viz., 
V m =Fx m , is left to expand. Assuming no device adopted 
(such as injecting hot-water spray) for preventing the cooling 
and rapid decrease of tension during expansion, the latter is 
adiabatic, and hence the tension at any point P between M 
and JV will be 

*=W— ) § - • [see §478; V=Fx]; . . (a) 

.'. Work of expansion 

= [~V= fF(p -Pn)dx = Ffpdn - Fp n (x n - a> m ), (2) 

and is represented by the area MPNL. 

But £pfa=p^£*-*fo=- 2 ^ xJ [Q-J~ y*] ; 

i.e.,^^ = 2i5 ?A [l-(J] i ]. . . (3) 
j$>w cutetitute (3) in (2) and then add (2) to (1), noting that 



COMPRESSED-AIR ENGINE. 



633 



F(p m -p n )x m - Fp n {x n - x m ) = Fp m xJl - °^ 

i_ x m p n 

which furthermore, since n and m are adiabatically related 
[see (a)], can be reduced to 

and we have finally : 

Toted work on piston- \ _ w _ Q w r, fx m \n , , 

raJ^ stroke \ ~ w - ^^VP* L 1 ~ ^T J J' • W 

But i^? m = F w , and the adiabatic relation holds good, 

(ff) = © ; Le -' © = (;£) ; 

therefore we may also write 

W=ZV m pJl-(^f\; .... (5) 

in which F TO = the volume which the mass of air used per 
stroke occupies in the state m, i.e., in the reservoir, where the 
tension is^> m and the absolute temperature = T m . 

To find the work done per pound of air used (or other unit 
of weight), we must divide W by the weight G = V m y m of 
the air used per stroke, remembering (eq. (13), § 472) that 



V m y m = [ V m p mVo T -] - (T m p ). 



Work per unit of weight of\_ c ,tp p 
air used in adiabatic working ) ~ m y j 






■ (6) 



The back-pressure p n =p a = one atmosphere. 
In (6) y = .0807 lbs. per cub. foot, p = 14.7 lbs. per sq. 
inch, and T ^ 273° Abs, Cent, or 492° Abs. Fahr. 



634 



MECHANICS OF ENGINEERING. 



It is noticeable in (6) that for given tensions p m and p n , the 
work per unit of weight of air used is proportional to the ab- 
solute temperature T m of the reservoir. The temperature T n 
to which the air has cooled at the end of the stroke is obtained 
as in Example 2, § 478, and may be far below freezing-point 
unless T m is very high or the ratio of expansion, x m : x n , large. 

Example. — Let the cylinder of a compressed-air engine have 
a section of F = 108 sq. in. and a stroke x n = 15 inches. The 
compressed air entering the cylinder is at a tension of 2 atmos. 
(i.e., p m = 29.4 lbs. per sq. in., and p n ^-p m — J)? ari( i a * a 
temperature of 27° Cent, (i.e., T m = 300° Abs. Cent.). Ke- 
quired the proper point of cut-off, or x m = ? , in order that the 
tension may fall to one atmosphere at the end of the stroke ; 
also the work per stroke, and the work per pound of air. Use 
the foot, pound, and second. 

From eq. (a), above, we have 



x~, 



=x n (^f= 1.25/ / j = 0.7875 ft. = 9.45 inches, 



and hence the volume of air in state m, used per stroke [eq. 
(5)] is 

V m = Fx m = Hf X 0.7875 = 0.5906 cubic feet ; 
while the work per stroke is 

W= 3 X 0.5906 X 29.4 X 144 X [1 - (J)* ] = 1545 ft. lbs., 
and the work obtained from each pound of air, eq. (6), 



■ OyOQAy 14.7X 144 

- 3 X 300X 0.0807 X 273 X 



[}-Vl> 



17810 



ft. lbs. per pound of air used. 

The temperature to which the air has cooted at cne ena 
stroke [eq. (2), § 478] is 



COMPRESSED-AIR ENGINE. 635 

T n = T m a 3 /—- = 300 X VT= 300 X .794 = 238° Abs. C. ; 

V Pm 

i.e., — 35° Centigrade. 

482. Remarks on the Preceding. — This low temperature is 
objectionable, causing, as it does, the formation and gradual 
accumulation of snow, from the watery vapor usually found 
in small quantities in the air, and the ultimate blocking of the 
ports. By giving a high value to T m , however, i.e., by heat- 
ing the reservoir, T n will be correspondingly higher, and also 
the work pei' pound of air, eq. (6). If the cylinder be encased 
in a " jacket" of hot water, or if spray of hot water be injected 
behind the piston during expansion, the temperature may be 
maintained nearly constant, in which event Mariotte's law will 
hold for the expansion, and more work will be obtained per 
pound of air ; but the point of cut-off must be differently 
placed. Thus if, in eq. (4), § 479, we make the back-pressure, 
which = (Fa -=- Fl)p b , equal to the value to which the air- 
pressure has fallen at the end of the stroke by Mariotte's law, 
we have 

W Zoierm! r ex^Zt k } = Fa ^ lo S" Q= V >P> h S- (j)> « 
and hence 

Work per unit of weight of air, \ __ m Po i (l\ / 9 \ 
with isothermal expansion j — m y T °^' e \aP ' 

Applying these equations to the data of the example, we 
obtain 

Work per unit of weight of air with iso- \ _ nq rp p Q 
thermal expansion \ ~~ ' m y~T ' 



whereas, with adiabatic expansion, work ) _ fi9 
per unit of weight of air is only } ~ 



T p 

r A 



636 



MECHANICS OF ENGINEEKING. 



Y 


n' j>j 


h 






H i 


A. i 

j, i 




X 




1 i 
K X nT ^ 




» 






V.V.-E 

























483. Double-acting Air-compressor, with Adiabatic Compres- 
sion. — This is the converse of § 481. In Fig. 530 we have the 
piston moving from right to left, compressing a mass of air 
which at the beginning of the stroke fills the cylinder. This is 

brought about by means of an external 
motor (steam-engine or turbine, e.g.) 
which exerts a thrust or pull along the 
piston-rod, enabling it with the help 
of the atmospheric pressure of the 
fresh supply of air flowing in behind 
it, to first compress a cylinder-full of 
air to the tension of the compressed 
air in the reservoir, and then, the 
port or valve opening at this stage, 
to force or deliver it into the reservoir. 
Let the temperature and tension of the 
cylinder-full of fresh air be T ni and 
jp nx , and the tension in the reservoir be p nil . Suppose the 
compression adiabatic. As the piston passes from E toward 
the left, the air on the left has no escape and is compressed, its 
tension and temperature increasing adiabatically until it reaches 
a value p mi = that in reservoir, at which instant, the piston 
being at some point D, a valve opens and the further progress 
of the piston simply transfers the compressed air into the re- 
servoir without further increasing its tension. Throughout 
the whole stroke the piston-rod has the help of one atmosphere 
pressure on the right face, since a new supply of air is entering 
on the right to be compressed in its turn on the return stroke. 
The work done from Eto D may be called the work of com- 
pression / that from D to 0, the work of delivery. 

[Since, here, dx and dW (or increment of work) have con- 
trary signs, we introduce the negative sign as shown.] 



Fig. 530. 



The work of compression = —J^F{p —p ni )dx. 



(lc) 



The work of delivery = —J D E{p mi —p ni 



)dx. 



a$ 



AIR-COMPRESSOR. 637 

In these equations only^ and x are variables. In the sum- 
mation indicated in (lc) p changes adiabatically ; in (Id) ]? is 
constant =p mj as now written. 

In the adiabatic compression the air passes from the state n x 
to the state m x (see N x and M 1 in figure). 

The summations in these equations being of the same form 
as those in equations (1) and (2) of § 481, but with limits in- 
verted, we may write immediately, 

Work per stroke = W= 3 F w j? TOl [l -f^-V . . (2) 



and 



Pm, 

r 



Workper unit of weight ) ^ p \~-\ _( P n i 
of air compressed \ ~ mi y j 1 |_ \p^ 



(3) 



The value of T mi , at the immediate end of the sudden com- 
pression, by eq. (2) of § (478), is 



-=<M- < 4 > 



T 



The temperature of the reservoir being T m , as in § 481 
(usually much less than T mi ), the compressed air entering it 
cools down gradually to that temperature, T m , contracting in 
volume correspondingly since it remains at the same tension 
f mi . The mechanical equivalent of this heat is lost. 

Let us now inquire what is the efficiency of the combination 
of air-compressor and compressed-air engine, the former sup- 
plying air for the latter, both working adiabatically, assuming 
that no tension is lost by the compressed air in passing along 
the reservoir between, i.e., that p mi = p m . Also assume (as 
already implied, in fact) that p Ul =p n = one atmos., and that 
the temperature, T ni , of the air entering the compressor cyl- 
inder is equal to that, T m , of the reservoir and transmission- 
pipe. 

To do this we need only find the ratio of the amount of 
work obtained from one pound (or other unit of weight) in the 
compressed-air engine to the amount spent in compressing onr 
jjound of air in the compressor. Calling this ratio 77, the 



638 



MECHANICS OF ENGINEERING. 



efficiency, and dividing eq. (6) of § 481 by eq. (3) of this para- 
graph, we have, with substitutions just mentioned, 



Abs. temp, of outer fi 



ree air 



T 



m x 



J Abs. temp, of air at end ' 
} of sudden compression, 



(5) 



or, substituting from eq. (4), and remembering that T ni = T m , 
we have also 

"=(£>' « 

also, since 






we may write 



_ T n _ Ab. tem. air leaving eng. cyl. ,~ 

T m Ab. tem. outer free air. 

For practical details of the construction and working of 
engines and compressors, and the actual efficiency realized, the 
student may consult special works, as they lie somewhat be- 
yond the scope of the present work. 

Example 1. — In the example of § 445, the ratio of p m to p n 
was = i . Hence, if compressed air is supplied to the reser- 
voir under above conditions, the efficiency of the system is, 
from eq. (6), rj — V i = 0.794, about 80 per cent. 

V 1 

Example 2. — If the ratio of the tensions is as small as — = — , 

the efficiency would be only (-$-)* = 0.55 ; i.e., 45 per cent of 
the energy spent in the compressor is lost in heat. 

Example 3. — What horse-power is required in a blowing 
engine to furnish 10 lbs. of air per minute at a pressure of 
4 atmos., with adiabatic compression, the air being received 
by the compressor at one atmosphere tension and 27° Cent. 



(ft.-lb.-sec. system) 
have, from eq. (4), 



Since 27° C. = 300° Abs. C. = T Wl , we 



w 2 > 



T ni = 300 ft)* = 477° Abs. Cent. ; 
and hence, eq. (3), 



HOT-AIH ENGINES. 639 

The work per pound of air = 3 X 477 0807 x 27 3 [} ~" (j J 

= 50870 ft. lbs. per pound of air. Hence 10 lbs. of air will 
require 508700 ft. lbs. of work ; and if this is done every min- 
ute we have the req. H. P. = WW- = 15 - 4 H - P - 

Note. — If the compression could be made isothermal, an 
approximation to which is obtained by injecting a spray of 
cold water, we would have, from eqs. (1) and (2) of § 482 : 

Workper) T P. ^„ ( Pm x \ 300 X 14.7 X 144 ^ QQA 

II. at hSfeM^r .0807X273 X L386 

s= 39950 ft. lbs. per lb., and the corresponding H. P. = 12.1 ; 
a, saving of about 25 per cent, compared with the former. 
The difference was employed in heating the air in the air-com- 
pressor with adiabatic compression, and was lost when that 
extra heat was dissipated in the reservoir as the air cooled 
again. This difference is easily shown graphically by compar- 
ing in the same diagram the areas representing the work done 
in the two cases. 

484. Hot-air Engines. — Since we have seen that the tension 
of air and other gases can be increased by heating, if the vol- 
ume be kept the same, a mass of air thus treated can after- 
wards be allowed to expand in a working cylinder, and thus 
become a means of converting heat into work. In Stirling's 
hot-air engine a definite confined mass of air.is used indefinitely 
without loss (except that occasional small supplies are needed 
to make up for leakage), and is alternately heated and cooled. 
A displacement-plunger, or piston, fitting loosely in a bell-like 
chamber, is so connected with the piston of the working 
cylinder and the fly-wheel, that its forward stroke is made 
while the other piston waits at the beginning of its stroke. 
In this motion the plunger causes the confined air to pass in a 
thin sheet over the top and sides of the furnace dome, thus 
greatly increasing its tension. The air then expands behind 
the working piston with falling tension and temperature, and, 



640 MECHANICS OF ENGINEEKING. 

while that piston pauses at the end of its forward stroke, is. 
again shifted in position, though without change of volume, 
by the return stroke of the plunger, in such a way as to pass 
through a coil of pipes in which cold water is flowing. This 
reduces both its temperature and tension, and hence its resist- 
ance to the piston on the return stroke is at first less than at- 
mospheric, but is gradually increased by the compression. 
This cycle of changes is repeated indefinitely, and is easily 
traced on a diagram like that in Fig. 528, and computations 
made accordingly. 

A special invention of Stirling's is the " regenerator" or box 
filled with numerous sheets of wire gauze, in its passage 
through which the working air, after expansion, deposits some 
of its heat, which it re-absorbs to some extent when, after 
further cooling in the " refrigerator" or pipe coil and com- 
pression by the return stroke of the piston, it is made to pass 
backward through the regenerator to be further heated by the 
furnace in readiness for a forward stroke. This feature, how- 
ever, has not realized all the expectations of its inventor and 
improvers, as to economy of heat and fuel. 

In Ericssovus hot-air engine, of more recent date, the dis- 
placement-plunger fits its cylinder air-tight, but valves can be 
opened through its edges when moving in one direction, thus 
causing it to act temporarily as a loose plunger, or shifter. 
The two pistons move simultaneously in the same direction in 
the same cylinder, but through different lengths of stroke, so 
that the space between them is alternately enlarged and con- 
tracted. The working piston also has valves opening through 
it for receiving a fresh supply of air into the space between 
the two pistons. During the forward stroke a fresh instal- 
ment from the outer air enters through the working piston into 
the space between it and the other, whose valves are now 
closed and which is now expelling from its further face, 
through proper valves, the air used in the preceding stroke ; 
no work is done in this stroke. On the return stroke this 
fresh supply of air is free to expand behind the now retreating 
working piston, while its tension is greatly increased by its 
being shifted (at least a large portion of it) over the furnace 



GAS-ENGINES. 641 

dome through the valves (now open) of the plunger piston, by 
the motion of the latter, which now acts as a loose plunger. 
The engine is therefore only single-acting, no work berng done 
in each forward stroke. For further details, see Goodeve's 
and Rankine's works on the steam-engine ; also the article 
" Hot-air Engine" in Johnson's Cyclopaedia by Pres. Barnard, 
and Rontgen's Thermodynamics. 

485. Brayton's Petroleum-engine. — Although a more recent 
invention than the gas-engines to be mentioned presently, this 
motor is more closely related to hot-air engines than the latter. 
By a slow combustion of petroleum vapor the gaseous products 
of combustion, while under considerable tension, are enabled 
to follow up a piston with a sustained pressure, being left to 
expand through the latter part of the stroke. Thus we have 
the furnace and working cylinder combined in one. The 
gradual combustion is accomplished by making use of the 
principle of the Davy safety-lamp that flame will not spread 
through layers of wire gauze of proper fineness. 

486. Gas-engines. — We again have the furnace and working 
cylinder in one in a " gas-engine" where illuminating gas and 
atmospheric air are introduced into the working cylinder in 
proper proportions (about ten parts of air to one of gas, by 
weight) to form an explosive mixture of more or less violence and 
exploded at a certain point of the stroke, causing a very sudden 
rise of temperature and tension, after which the mass expands 
behind the piston with falling pressure. On the return stroke 
the products of combustion are expelled, and no work done, 
these engines being single-acting. In some forms the mixture 
is compressed before explosion, since it has been found that 
under this treatment a mixture containing a larger proportion 
of air to gas can be made to ignite, and that then the resulting 
pressure is more gradual and sustained, like that of steam or of 
the mixture in the Brayton engine. That is, the effect is 
analogous to that of " slow-burning powder" in a gun. 

In the " Otto Silent Gas-engine" the explosion occurs only 
every fourth stroke, and one side of the piston is always open 



642 MECHANICS OF ENGINEEEING. 

to the air. The action on the other side of the piston is as 
follows : (1) In the forward stroke a fresh supply of explosive 
mixture is drawn into the cylinder at one atmosphere tension. 
(2) The next (backward) stroke compresses the mixture into 
about one fourth of its original bulk, this operation occurring 
at the expense of the kinetic energy of the fly-wheel. (3) The 
mixture is ignited, the pressure rises to 8 or 10 atmospheres, 
and work is done on the piston through the next (forward) 
stroke, the tension of the products of combustion having 
fallen to about two atmospheres at the end of the stroke. 
(4) In the next (backward) stroke the products of combus- 
tion are expelled and no work is done. 

The Atkinson "Cycle Gas-engine," an English invention of 
recent date (see the London Engineer for May 1887 ; pp. 361 
and 380) also makes an explosion every fourth stroke, but the 
link work connecting the piston and fly-wheel is of such de- 
sign that the latter makes but one revolution during the four 
strokes. Also the length of the expansion or working stroke 
is greater than that of the compression stroke and the products 
of combustion are completely expelled. Consequently the effi- 
ciency of this motor is at present greater than that of any other 
gas-engine. See § 487. 

One of the most simple gas-engines is made by the Economic 
Motor Company of New York. The piston has no packing, 
being a long plunger ground to fit the cylinder accurately and 
kept well lubricated. As with most gas-engines the cylinder is 
encased in a water-jacket to prevent excessive heating of the 
working parts and consequent decomposition of the lubricant. 

For further details on these motors, see Rankine's " Steam- 
engine," Clark's " Gas-engines " in Yan Nostrand's Science 
Series, and article " Gas-engine" in Johnson's Cyclopaedia ; also 
Prof. Thurston's report on Mechanical Engineering at the 
Yienna Exhibition of 1873, and proceedings of the " Society 
of Engineers" (England) for 1881. 

487. Efficiency of Heat-engines. — According to the mechan- 
ical theory of heat, the combustion of one pound of coal, pro- 



GAS-ENGINES. 643 

during, as it does, about 14,000 heat-uoits (British Thermal 
units ; see § 149, Mechanics) should furnish 

14,000 X 772 = 10,808,000 ft. lbs. of work, 

if entirely converted into work. Let us see how nearly this is 
accomplished in the performance of the most recent and 
economical marine engines of the present day, viz., the triple 
expansion engines of some Atlantic steamers, which are claimed 
to have consumed per hour only 1.25 lbs. of coal for each 
measured (" indicated ") horse-power of effective work done in 
their cylinders. The work-equivalent of 1.25 lbs. of coal per 
hour is 

1.25 X 14,000 X 772 = 13,510,000 ft. lbs. per hour ; 

while the actual work per hour implied in " one H. P. per 
hour" is 

33000 X 60 = 1,980,000 ft. lbs. per hour. 

That is, the engines utilize only one seventh of the heat of com- 
bustion of the fuel. 

According to Prof. Thurston, this is a rather extravagant 
claim (1.25), the actual consumption having probably been 1.4 
lbs. of coal per H. P. per hour. 

The ordinary compound marine engine is stated to use as 
little as 2.00 lbs. per hour for each H. P. 

Most of the heat not utilized is dissipated in the condenser. 

Similarly, the water-jacket, a necessary evil in the operation 
of the gas-engine, is a source of great loss of heat and work. 
Still, Mr. Wm. Anderson in his recent work, " Conversion of 
Heat into Work" (London, 1887), mentions a motor of this 
class as having converted into work i of the heat of combus- 
tion [an Otto " Silent Gas-engine," tested at the Stevens 
Institute, Hoboken, N. J., in 1883] ; while Prof. Unwin found 
the Atkinson engine (see last paragraph) capable of returning 
(in the cylinder) fully -§- of the heat-equivalent of the gas con- 
sumed. [This latter result was confirmed in Philadelphia in 
Jan. 1889 by Prof. Barr, under direction of Prof. Thurston.] 



644 MECHANICS OF ENGINEEKING. 

488. Duty of Pumping-engines. — Another way (often used 
in speaking of the performance of pum ping-engines j of ex- 
pressing the degree of economy attained in the use of fuel by 
the combined furnace, boiler, and engine is to give the num- 
ber of foot-pounds of work obtained from each 100 lbs. of coal 
consumed in the furnace, calling it the " duty" of the engine. 

For example, by a duty of 99,000,000 ft. lbs. it is meant that 
from each pound of coal 990,000 ft. lbs. of work are obtained. 
From this we gather that, since one horse-power consists of 
33,000 X 60 = 1,980,000 ft. lbs. per hour, the engine men- 
tioned must use each hour 

1,980,000 -f- 990,000 = 2 lbs. of coal for each H. P. developed j 

which is as low a figure as that attained by the marine engines 
last quoted. 

489. Buoyant Effort of the Atmosphere. — In the case of a 
body of large bulk but of small specific gravity the buoyant 
effort of the air (due to the same cause as that of water, see 
§ 456) becomes quite appreciable, and may sometimes be 
greater than the weight of the body. This buoyant effort is 
equal to the weight of air displaced, i.e., = Vy, where V is 
the volume of air displaced, and y its heaviness. 

If G x — total weight of the body producing the displace- 
ment, the resultant vertical force is 

P=G l -Vy, (1) 

and for equilibrium, or suspension in the air, we must have 
P = 0, i.e., 

G^Vy . (2) 

We may therefore find approximately the elevation where 
a given balloon will cease to ascend, by determining the heavi- 
ness y of the air at that elevation from eq. (2) ; then, know- 
ing approximately the temperature of the air at that elevation, 
we may compute its tension p [eq. (13), § 472], and finally, 
from eqs. (3), (4), or (5) of § 477, obtain the altitude required. 

Example. — The car and other solid parts of a balloon weigh 



.jLil# 



BALLOONS. 645 

400 lbs., arid the bag contains 12,000 cub. feet of illuminating 
gas weighing 0.030 lb. per cub. foot at a tension of one at- 
mosphere and temperature of 15° Cent., so that its total 
weight =12,000 X 0.030 = 360 lbs. 

Hence G x — 760 lbs. We may also write with sufficient 
accuracy : Whole volume of displacement = V = 12,000 cub. ft. 

As the balloon ascends the exterior pressure diminishes, and 

the confined gas tends to expand and so in- _m__ q_ _^_. m 

crease the volume of displacement V; but .'•'•.'•.':'••'•;.'•.;'••'' 
this we shall suppose prevented by the *.'*•/.'.• .\\v."i •'•'. 
strength of the envelope. At the surface . ••/; ;..*'. 7. ".' j"-'. : 
of the ground (station n of Fig. 531; see ;."•.';: \\\:} \i--T 
also Fig. 526) let the barometer read 29.6 V.: •.'••' :•//.'•. 
inches and the temperature be 15° Cent. /f/7W^p^= 
Then T n — 288° Abs. Cent., and the heavi- fig. 531. 

ness of the air at n is 

= .0807 X 273 afcfl X 14.7 
Yn ~ 14.7 ' 288 

(.^f) = .080 7 x|g.^=.0r 54 l b , per cub. f t. 

At the unknown height A, where the balloon is to come to 
rest, i.e., at M, G x must = Vy [eq. (2)] ; therefore 

G 1 760 lbs. A „ OQ ,, , , 

y ~=v = ipooTu-Ot = - 0633 lbs - P er cub - ft ; 

and if the temperature at M be estimated to be 5° Cent, (or 
T m = 278° Abs. Cent.) (on a calm day the temperature de- 
creases about 1° Cent, for each 500 ft. of ascent), we shall 

have, from Jfy- = Jfy.,. 

Pn = rnTn = ^m 288 

Pm YmT m .0633*278 ' ' 
and hence, from eq. (5), § 477, with i(T m + T n ) put for T n , 
h = 26213 X fff X 2.30258 X log. 10 1.206 = 5088 ft. 



CHAPTEE VI. 



HYDRODYNAMICS BEGUN— STEADY FLOW OF LIQUIDS 
THROUGH PIPES AND ORIFICES. 



489a. The subject of Water in Motion presents one of the 
most unsatisfactory branches of Applied Mechanics, from a 
mathematical stand-point. The internal eddies, cross-currents, 
and general intricacy of motion of the particles among each 
other, occurring in a pipe transmitting a fluid, are almost en- 
tirely defiant of mathematical expression, though the flow of 
water through a circular orifice in a thin plate into the air pre- 
sents a simpler case, where the conception of " stream lines" is 
probably quite close to the truth. In most practical cases we 
are forced to adopt as a basis for mathematical investigation 
the simple assumption that the particles move side by side in 
such a way that those which at any instant form a lamina 
or thin sheet, 1 to the axis of the pipe or orifice, remain 
together as a lamina during the further stages of the flow. 
This is the Hypothesis of Flow in Plane Layers, or Laminated 
Flow. Experiment is then relied on to make good the discre- 
pancies between the indications of the formulae resulting from 
this theory and the actual results of practice ; so that the science 
of Hydrodynamics is largely one of coefficients determined by 
experiment. 

490. Experimental Phenomena of a " Steady Flow." — As pre- 
liminary to the analysis on which the formulae of this chapter 
are based, and to acquire familiarity with the quantities involved, 
it will be advantageous to study the phenomena of the appara- 
tus represented in Fig. 532. A large tank or reservoir IZCis 
connected with another, DE, at a lower level, by means of a 
rigid pipe opening under the water-level in each tank. This 



STEADY FLOW OF A LIQUID. 



647 



pipe has no sharp curves or bends, is of various sectional areas 
at different parts, the changes of section being very gradual, 
and the highest point jY" a not being more than 30 ft. higher 
than BC, the surface-level of the upper tank. Let both tanks 




Fig. 532. 



be filled with water (or other liquid), which will also rise to H 
and to iT in the pipe. Stop the ends L and iV 4 of the pipe, 
and through M, a stop-cock in the highest curve, pour in water 
to fill the remainder of the pipe ; then, closing M, unstop L 
and iT 4 . 

If the dimensions are not extreme (and subsequent formulae 
will furnish the means of testing such points) the water will 
now begin to flow from the upper tank into the lower, and 
all parts of the pipe will continue full of water as the flow 
goes on. 

Further, suppose the upper tank so large that its stirf ace- 
level sinks very slowly ; or that an influx at A continually 
makes good the efflux at E\ then the flow is said to be a Steady 
Flow; or, a state of permanency is said to exist; i.e., the cir- 
cumstances of the flow at each section of the pipe are per- 
manent, or steady. 



648 MECHANICS OF ENGINEERING. 

By measuring the volume, V, of water discharged at E in a 
time t, we obtain the volume of flow per unit of time, viz., 

«=•?' (1) 

while the weight of flow per unit of time is 

= Qr, (2) 

where y = heaviness (§ 7) of the liquid concerned. 

Water being incompressible and the pipe rigid, it follows 
that the same volume of water per unit of time must be pass- 
ing at each cross-section of the pipe. But this is equal to the 
volume of a prism of water having F, the area of the section, 
as a base, and, as an altitude, the mean velocity = v with which 
the liquid particles pass through the section. Hence for any 
section we have 

Q = Fv = Q0Q ^nt = F^F, % ,et e .\f^ u ^\, . (3) 

in which the subscripts refer to different sections. If the flow 
were unsteady, e.g., if the level BC were sinking, this would 
be true for a definite instant of time ; but when steady, we 
see that it is permanently true; e.g.,F 1 v 1 at any instant == F^o % 
at the same or any other instant, subsequent or previous. In 
other words, in a steady flow the velocity at a given section 
remains unchanged with lapse of time. 

[N.B. We here assume for simplicity that the different 
particles of water passing simultaneously through a given sec- 
tion (i.e., abreast of each other) have equal velocities, viz., the 
velocity which all other particles will assume on reaching this 
section. Strictly, however, the particles at the sides are some- 
what retarded by friction on the surface of the pipe. This as- 
sumption is called the Assumption of Parallel Flow, or Flow 
in Plane Layers, or Laminated Flow.] 

Let us suppose Q to have been found as already prescribed. 
We may then, knowing the internal sectional areas at different 
parts of the pipe, iV^ , iV^, etc., compute the velocities 



STEADY FLOW OF LIQUIDS. 649 

%=Q+J[, v,= Q-~F„ etc., 

•which the water must have in passing those sections, respec- 
tively. It is thus seen that the velocity at any section has no 
direct connection with the height or depth of the section from 
the plane, BO, of the upper reservoir surface. The fraction 

v 2 

-— will be called the height due to the velocity, v, or simply 

the velocity -head, for convenience. 

Next, as to the value of the internal fluid pressure, p, per 
unit-area (in the water itself and against the side or wall of 
pipe) at different sections of the pipe. If the end N± of the 
pipe were stopped, the problem would be one in Hydrostatics, 
and the pressure against the side of the pipe at N x (also at JV 3 
on same level) would be simply 

measured by a water column of height 

in which p a = one atmosphere, and b = 34 ft. = height of an 
ideal water barometer, and y = 62.5 lbs. per cubic foot ; and 
this would be shown experimentally by screwing into the side 
of the pipe at N~ l a small tube open at both ends ; the water 
would rise in it to the level BO. That is, a column of water 
of height = h x would be sustained in it, which indicates that 
the internal pressure at N 1 corresponds to an ideal water col' 
umn of a height 

But when a steady flow is proceeding, the case being now one 
of Hydrodynamics, we find the column of water sustained at 
rest in the small tube (called an open piezometer) Nfi has a 
height y 1 , less than h 1 , and hence the internal fluid pressure is 



650 MECHANICS OF ENGINEEKING. 

less than it was when there was no flow. This pressure being 
called^ , the ideal water column measuring it has a height 

f- = t+y> (±> 

at JV 1 , and will be called the pressure-head at the section re- 
ferred to. We also find experimentally that while the flow is 
steady the piezometer-height y (and therefore the pressure- 
head h -\- y) at any section remains unchanged with lapse of 
time, as a characteristic of a steady flow. 

[For correct indications, the extremity of the piezometer 
should have its edges flush with the inner face of the pipe 
wall, where it is inserted.] 

At iT, , although at the same level as i\^ , we find, on in- 
serting an open piezometer, W, that with F 3 = F x (and there- 
fore with v z — Vj) y z is a little less than y 1 ; while if F 3 < F x 
(so that i> 3 > ^ x ), y z is not only less than y 1 , but the dif- 
ference is greater than before. We have therefore found 
experimentally that, in a general way, when water is flowing 
in a pipe it presses less against the side of the pipe than it did 
before the flow was permitted, or (what amounts to the same 
thing) the pressure between the transverse laminse is less than 
the hydrostatic pressure would be. 

In the portion HNfi of the pipe we find the pressure less 
than one atmosphere, and consequently a manometer register- 
ing pressures from zero upward (and not simply the excess 
over one atmosphere, like the Bourdon steam-gauge and the 
open piezometer just mentioned) must be employed. At iT, , 
e.g., we find the pressure 

= i atmos., i.e., Si = 17 ft. 
Y 

Even below the level BC,hj making the sections quite nar- 
row (and consequently the velocities great) the pressure may be 
made less than one atmosphere. At the surface BC the, pres- 
sure is of course just one atmosphere, while that in the jet at 
iT 4 , entering the right-hand tank under water, is necessarily 

p t = 1 atmos. -f- press, due to col. h! of water practically at rest; 






STEADY FLOW OF LIQUIDS. 651 

i.e., ±-± = pressure-head at J¥ A = b -f- h f , 

(whereas if N^ were stopped by a diaphragm, the pressure- 
head just on the right of the diaphragm would be b -\- h\ and 
that on the left b -f- h i .) 

Similarly, when a jet enters the atmosphere in parallel fila- 
ments its particles are under a pressure of one atmosphere, i.e., 
their pressure-head = b = 34 ft. (for water) ; for the air im- 
mediately around the jet may be considered as a pipe between 
which and the water is exerted a pressure of one atmosphere. 

491. Recapitulation and Examples. — We have found experi- 
mentally, then, that in a steady flow of liquid through a rigid 
pipe there is at each section of the pipe a definite velocity and 
pressure which all the liquid particles assume on reaching that 
section ; in other words, at each section of the pipe the liquid 
velocity and pressure remain constant with progress of time. 

Example 1. — If in Fig. 532, the flow having become steady, 
the volume of water flowing in 3 minutes is found on meas- 
urement to be 134 cub. feet, the volume per second is, from 
eq. (1), § 490, 

Q = -is*. == 0.744 cub. ft. per second. 

Example 2. — If the flow in 2 min. 20 sec. is 386.4 lbs., the 
volume of flow per second is [ft., lb., sec. ; eqs. (1) and (2)] 

n V G . 386.4 1 „_,., , , 

Q — — = r- 1 — — — - . — — — 0.0441 cub. ft. per sec. 

^ t y 62.5 140 v 

Example 3. — In Fig. 532 the height of the open piezometer 
at J¥ l is y 1 = 9 feet ; what is the internal fluid pressure ? 
[Use the inch, lb., and sec] The internal pressure is 

p i =p a J r y i y = 14.7 + 108 X fM = 18 - 6 lbs - P er 8q- incn - 

The pressure on the outside of the pipe is, of course, one at- 
mosphere, so that the resultant bursting pressure at that point 
(iTj) is 3.9 lbs. per sq. in. 

Example 4. — The volume of flow per second being .0441 



652 MECHANICS OF ENGINEEKING. 

cub. ft. per sec, as in Example 1, required the velocity at a 
section of the (circular) pipe where the diameter is 2 inches. 
[Use ft., lb., and sec] 

Q 0.0441 _ AO ,. 

while at another section of the pipe where the diameter is four 
inches (double the former) and the sectional area, F, is there- 
fore four times as great, the velocity is \ of 2.02 = 0.505 ft. 
per sec. 

492. Bernoulli's Theorem for Steady Flow ; without Friction. — 

If the pipe is comparatively short, without sudden bends, 
elbows, or abrupt changes of cross-section, the effect of friction 
of the liquid particles against the sides of the pipe and against 
each other (as when eddies are produced, disturbing the paral- 
lelism of flow) is small, and will be neglected in the present 
analysis, whose chief object is to establish a formula for steady 
flow through a short pipe and through orifices. 

An assumption, now to be made, of flow in plane layers, or 
laminated flow, i.e., flow in laminae ~| to the axis of the pipe 
at every point, may be thus stated : (see Fig. 533, which shows 
P/Pe a steady flow proceeding, through a 

pipe CD of indefinite extent.) All the 
liquid particles which at any instant 
form a small lamina, or sheet, as AB, 
"I to axis of pipe, keep company as a 
la?nina throughout the whole flow. 
fig. 533. The thickness, ds\ of this lamina re- 

mains constant so long as the pipe is of constant cross-section, 
but shortens up (as at C) on passing through a larger section, 
and lengthens out (as at D) in a part of the pipe where the 
section is smaller (i.e., the sectional area, F, is smaller). The 
mass of such a lamina is Fds'y -f- g [§ 55], its velocity at any 
section will be called v (pertaining to that point of the pipe's 
axis), the pressure of the lamina just behind it is Fp, upon the 
rear face, while the resistance (at the same instant) offered by 
its neighbor just ahead is F(p -f- dp) on the front face ; also 




Bernoulli's theorem. 



its weight is the vertical force Fds'y. Fig. 534 shows, as a 
free body, the lamina which at 
any instant is passing a point 
A of the pipe's axis, where the 
velocity is v and pressure jp. 

Note well the forces acting ; 
the pressures of the pipe wall 
on the edges of the lamina have 
no components in the direction 
of v, for the wall is considered 
smooth, i.e., those pressures are 
"1 to wall ; in other words, no 
friction is considered. To this free body apply eq. (7) of § 74, 
for any instant of any curvilinear motion of a material point 




> r dG=Fdsy 

Fig. 534. 



vdv = (tang, acceleration) X ds, . 



(i) 



in which ds = a small portion of the path, and is described in 
the time dt. Now the tang, accel. = ^"(tang. com pons, of the 
acting forces) -f- mass of lamina, i.e., 



tang. ace. 



_F P 



F(jp -f- dp) -f- Fyds' cos 
Fyds' -T- g 



(2) 



Now, Fig. 535, at a definite instant of time, conceive the 
volume of water in the pipe to be subdivided into a great 
number of laminae of equal mass (which implies equal volumes 











Fig. 535. 



in the case of a liquid, but not with gaseous fluids), and let the 
ds just mentioned for any one lamina be the distance from its 
centre to that of the one next ahead ; this mode of subdivision 



654 MECHANICS OF ENGINEEKING. 

makes the ds of any one lamina identical in value with its 
thickness ds', i.e., 

ds = ds' (3) 

We have also 

ds cos = — dz, or ds' cos = — dz ; . . (4) 

z being the height of the centre of a lamina above any con- 
venient horizontal datum plane. Substituting from (2), (3), 
and (4) in (1), we derive finally 

- vdv -\ — dp 4-dz = (5) 

9 Y 

The flow being steady, and the subdivision into laminae 
being of the nature just stated, each lamina in some small time 
dt moves into the position which at the beginning of dt was 
filled by the lamina next ahead, and acquires the same velocity, 
the same pressures on its faces, and the sam,e value of z, that 
the front lamina had at the beginning of dt. 

Hence, considering the simultaneous advance made by all 
the laminae in this same dt, we may write out an equation like 
(5) for each of the laminae between any two cross-sections n and 
m of the pipe, thus obtaining an infinite number of equations, 
from which by adding corresponding terms, i.e., by integra- 
tion, we obtain 

-J Vn vdv+-J Pn d2>+J z j3 = 0; ... (6) 

whence, performing the integrations and transposing, 

I'm \Pm\ 9 _ V n , Pn , „ ( Bernoulli's \ ,~ 

~^ + y + m ""2^" l ~7"' t ~ n ' '( Theorem [' ' V \ 

Denoting by Potential Head the vertical height of any section 
of the pipe above a convenient datum level, we may state 
Bernoulli's Theorem as follows : 

In steady flow without friction, the sum of the velocity- 
head, pressure-head, and potential head at any section of the 
pipe is a constant quantity, being equal to the sum of the cor- 
responding heads at any other section. 



APPLICATIONS OF BEKNOULLTS THEOREM. 



m 



It is noticeable that in eq. (7) each of the terms is a linear 
quantity, viz., a height, or head, either actual, such as z n and 
2 m , or ideal (all the others), and does not bring into account the 
absolute size of the pipe, nor even its relative dimensions (v m 
and v n , however, are connected by the equation of continuity 
F m v m = F n v n \ and contains no reference to the volume of 
water flowing per unit of time \_Q] or the shape of the pipe's 
axis. "When the pipe is of considerable length compared with 
its diameter the friction of the water on the sides of the pipe 
cannot be neglected (§ 512). 

It must be remembered that Bernoulli's Theorem does not 
hold unless the flow is steady^ i.e., unless each lamina, in com- 
ing into the position just vacated by the one next ahead (of 
equal mass), comes also into the exact conditions of velocity 
and pressure in which the other was when in that position. 

[N.B. This theorem can also be proved by applying to all 
the water particles between n and m, as a collection of small 
rigid bodies (water being incompressible) the theorem of Work 
and Energy for a collection of Rigid Bodies in § 142, eq. (xvi), 
taking the respective paths which they describe simultaneously 
in a single dt.~\ 

493. First Application of Bernoulli's Theorem without Friction. 
— Fig. 536 shows a large tank from which a vertical pipe of 
uniform section leads to another tank and dips below the sur- 
face of the water in the latter. Both surfaces are open to the 
air. The vessels and pipe being filled with 
water, and the lower end m of the pipe un- 
stopped, a steady flow is established almost 
immediately, the surface BC being very 
large compared with F, the area of the {uni- 
form) section of the pipe. 

Given F, and the heights h Q and A, re- 
quired the velocity v m of the jet at m and 
also the pressure, p n , at n (in pipe near en- 
trance of same), m is in the jet, just clear 
of the pipe, and practically in the water- 
level, AD. The velocity v m is unknown, Fl& ' 536 - 
but the pressure p m is practically =p a = one atmosphere, since 





"IB -..-. 


air: 


:-' : en 


'X ' 


!P^ 


-— 


j 




?==1 


( 

fo 
1 

1 

1 

ft 


* 

* 

1 

1 

A ^ 


lljif 

§ 
III 


n 

••'air*. 
DJ1 




iSSf 





656 



MECHANICS OF ENGINEERING. 



the pressure on the sides of the jet is necessarily the hydro- 
static pressure due to a slight depth below the surface AD. 



Press.-head at m is^ = ^ = b = 34 feet. 
Y Y 



(§423) 



Now apply Bernoulli's Theorem to sections m and n, taking 
a horizontal plane through m as a datum plane for potential 
heads, so that z n = h and z m = 0, and we have 



2g ^ T ^9^ Y 



(1) 



But, assuming that the section of the pipe is filled at every 
point, we must have 

v m . = v„ 



'm — w n 5 



for, in the equation of continuity 

if we put F m = F n , the pipe being of uniform section, we ob- 
tain v m = v n . Hence eq. (1) reduces to 



£z=zb-h=te&-h. 
Y 



(2) 



mm 



Sv 



HA 



Hence the pressure at n is less than one atmosphere, and if a 
small tube communicating with an air-tight receiver full of air 
were screwed into a small hole at n, the air in 
the receiver would gradually be drawn off until 
its tension had fallen to a value p n . [This is the 
principle of SprengePs air-pump, mercury, how- 
ever, being used instead of water, as for this 
heavy liquid b — only 30 inches.] 

If h is made > b for water, i.e. > 34 feet (or 
> 30 inches for mercury), p n would be negative 
from eq. (2), which is impossible, showing that 
the assumption of full pipe-sections is not borne 
out. In this case, h> b, only a portion, mn\ 
(in length somewhat less than b,) of the tube will be kept full 






U\ 



tr^vtfi^ 



Fig. 537. 



APPLICATIONS OF BERNOULLI'S THEOREM. 657 

during the flow (Fig. 537); while in the part Kn' vapor of 
water, of low tension corresponding to the temperature 
(§ 469), will surround an internal jet which does not fill the 
pipe. As for the value of v m , Bernoulli's Theorem, applied 
to BC and m, in Fig. 536, gives finally v m — V2gh . 

Example. — If h = 20 feet, Fig. 536, and the liquid is water, 
the pressure-head at n is (ft., lb., sec.) 

£» = b - h = 34/ - 20' = 14 ft., 
V 

and therefore 
p n = 14 X 62.5 = 875 lbs. per sq. ft. = 6.07 lbs. per sq. in. 

494. Second Application of Bernoulli's Theorem without Fric- 
tion. — Knowing by actual measurement the open piezometer 
height y n at the section n in _ 
Fig. 538 (so that the pressure- B 

head, — = b -f- y n , at n is 

y 

known) ; knowing also the 

vertical distance h n from n 

to fn, and the respective fig. 538. 

cross-sections F n and F m (F m being the sectional area of the 

jet, flowing into the air, so that ^ = l\ required the volume 

<af flow per sec; i.e., required Q, which 

= F n v n = F m v m (1) 

The pipe is short, with smooth curves, if any, and friction 
will therefore be neglected. From Bernoulli's Theorem [eq. 
(7), § 492J, taking m as a datum plane for potential heads, we 
have 

|! + 5 + = g + ( 2/n + J) + A n . ... (2) 
But from (1) we have 



~^ = — ~ 0/ II 


••.":".• .- # 


— — — "^ /^"^l^^ 


■'■}* '--h. 


DATUM 






c- /m^ 



658 MECHANICS OF ENGINEEKING. 

substituting which in (2) we obtain, solving for v m , 



„. = * *<*.+K> _ (S) 



\/'-(f)' ' 



and hence the volume per unit of time becomes known, viz., 

Q = F m v m . ......... (4) 

Note. — If the cross-section F m of the nozzle, or jet, is > F n , 
v m becomes imaginary (unless y n is negative (i.e., p n < one at- 
mos.), and numerically > h n ) ; in other words, the assigned 
cross-sections are not filled by the flow. 

Example. — If y n = 17 ft. (thus showing the internal fluid 
pressure at n to be p n = y(y n -f- b) = 1^ atmos.), h n = 10 ft., 
and the (round) pipe is 4 inches in diameter at n and 3 inches 
at the nozzle m, we have from (3) (using ft.-lb.-sec. system of 
units in which g = 32.2) 



-..- fax 38.8(17 + 10) = aAtt _ 

-OS]' 



si 



[N.B. Since F m -f- i^ is a ratio and therefore an abstract 
number, the use of the inch in the ratio will give the same 
result as that of the foot.] 

Hence, from (4), 

Q = F m v m = i*(A) a X 50.4 = 2.474 cub. ft. per sec. 

495. Orifices in Thin Plate.— Fig. 539. When efflux takes 
place through an orifice in a thin plate, i.e., a sharp-edged 
orifice in the plane wall of a tank, a contracted vein (or " vena 



ORIFICE IN THIN PLATE. 



659 



contracta") is formed, the filaments of water not becoming 
parallel until reaching a plane, m, 
parallel to the plane of vessel wall, 
which for circular orifices is at a dis- 
tance from, the interior plane of vessel 
wall equal to the radius of the circular 
aperture ; and not until reaching this 
plane does the internal fluid pres- 
sure become equal to that of the sur- 
rounding medium (atmosphere, here), 
i.e., surrounding the jet. We assume 






Fig. 539. 

that the width of the 
orifice is small compared with A, unless the vessel wall is 
horizontal. 

The area of the cross-section of the jet at m, called the con- 
tracted section, is found on measurement to be from .60 to .64 
of the area of the aperture with most orifices of ordinary 
shapes, even with widely different values of the area of aper- 
ture and of the height, or head, A, producing the flow. Call- 
ing this abstract number [.60 to .64] the Coefficient of Con- 
traction, and denoting it by C, we may write 



F m =CF, 



in which F — area of the orifice, and F m = that of the con- 
tracted section. C ranges from .60 to .64 with circular orifices, 
but may have lower values with some rectangular forms. (See 
table in § 503.) 

A lamina of particles of water is under atmospheric 
pressure at n (the free surface of the water in tank or reser- 
voir), while its velocity at n is practically zero, i.e. v n — 
(the surface at B being very large compared with the area of 
orifice). It experiences increasing pressure as it slowly de- 
scends until in the immediate neighborhood of the orifice, 
when its velocity is rapidly accelerated and pressure decreased, 
in accordance with Bernoulli's Theorem, and its shape length- 
ened out, until finally at m it forms a portion of a filament of 
a jet, its pressure is one atmosphere, and its velocity, = v m , 
we wish to determine. The course of this lamina we call a 



660 



MECHANICS OF ENGINEEKING. 



" stream-line" and Bernoulli's Theorem is applicable to it, 
just as ii it were enclosed in a frictionless pipe of the same 
form. Taking then a datum plane through the centre of m, 
we have 

^ = b, z m = 0, and v m = !; 

r 

while 

^ also = b, z n = h, and v n = 0. 

r 



Hence Bernoulli's Theorem gives 



2<? 



+ 5 + = + 5 + A; 



2<7 



h, 



(1) 



and 



v m = V2gh. 



That is, the velocity of the jet at m is theoretically the same as 
that acquired by a body falling freely in vacuo through a 
height =h= the " head of water." We should therefore ex- 
pect that if the jet were directly ver- 
tically upward, as at m, Fig. 540, 




a height 



2? 



Fig. 540. 



would be actually attained. [See 
§§ 52 and 53.] Experiment shows 
that the height of the jet (at m) 
does not materially differ from h if 
A is not > 6 or 8 feet. For h > 8 ft., however, the actual height 
reached is < A, the difference being not only absolutely but 
relatively greater as h is taken greater, since the resistance of 
the air is then more and more effective in depressing and 
breaking up the stream. (See § 578.) 

At m\ Fig. 540, we have a jet, under a head = h', directed 



OEIFICE IN THIN PLATE. 661 

at an angle a with the horizontal. Its form is a parabola 
(§ 81), and the theoretical height reached is h" = hi sin 2 a 
(§ 80). 

The jet from an orifice in thin plate is very limpid and clear. 

From eq. (1), we have theoretically 

v m = V%jh 

(an equation we shall always use for efflux into the air through 
orifices and short pipes in the plane wall of a large tank whose 
water-surface is very large compared with the orifice, and is 
open to the air), but experiment shows that for an " orifice in 
thin plate" this value is reduced about Sfo by friction at the 
edges, so that for ordinary practical purposes we may write 



v m = (pV2gh = 0.97 V2gh, .... (2) 

in which cp is called the coefficient of velocity. 

Hence the volume of flow, Q, per time-unit will be 



Q — F m v m = CF(p V2gh, on the average = 0.62FV 2gh. (3) 

It is to be understood that the flow is steady, and that the 
reservoir surface (very large) and the jet are both under at- 
mospheric pressure. 0(7 is called the coefficient of efflux. 

Example 1. — Fig. 539. Required the velocity of efflux, 
v m , at m, and the volume of the flow per second, Q, into the 
air, if h = 21 ft. 6 inches, the circular orifice being 2 in. in 
diam. ; take C = 0.64. [Ft., lb., and sec] 

From eq. (2), 



v m = 0.97 V2 X 32.3 X 21.5 = 36.1 ft. per sec. ; 
hence the discharge is 

Q = F m v m = 0.64 X j U^r) 2 X 36.1 = 0.504 cub. ft. per second. 

Example 2. — fWeisbach.] Under a head of 3.396 metres 
the velocity v m in the contracted section is found by measure- 



662 MECHANICS OF ENGINEERING. 

ments of the jet-curve to be 7.98 metres per sec, and the dis- 
charge proves to be 0.01825 cub. metres per sec. Required 
the coefficient of velocity (0) and that of contraction (<7), if 
the area of the orifice is 36.3 sq. centimetres. 

Use the metre-kilogram-second system of units, in which 
g = 9.81 met. per sq. second. 

From eq. (2), 

= -g^ = - 198 = = 0.978; 

V 2gh V2 X 9.81 X 3.396 

while from (3) we have 

c= Q_ = Q = -01825 =()631 

F4> V 2gfi Fv m -rihNSr X ^^ 

and C, being abstract numbers, are independent of the sys- 
tem of concrete units adopted. 

Note. — To find the velocity v m of the jet at the orifice by 
measurements of the jet-curve, as mentioned in Example 2, 
we may proceed as follows : Since we cannot very readily as- 
sure ourselves that the direction of the jet at the orifice is 
horizontal, we consider the angle a of the parabola (see Fig. 
93 and § 80) as unknown, and therefore have two unknowns 
to deal with, and obtain the necessary two equations by meas- 
uring the x and y (see page 84) of two points of the jet, re- 
membering that if we use the equation (3) of page 84 in its 
present form points of the jet below the orifice will have nega- 
tive ,y's. The substitution of these values #! , a? a , y t , and y z 
in equation (3) furnishes two equations between constants, in 
which only a and h are unknown. To eliminate a , for 

— - — we write 1 -f- tan 2 a , and taking x 2 = 2x 1 for coil- 
cos a 

venience, we finally obtain 



in which y x and y 2 are the vertical distances of the two points 



BOUNDED ORIFICE. 663 

chosen below the orifice ; that is, we have already made them 
negative in eq. (3) of page 84. The h of the preceding equa- 
tion simply denotes v m a 4- 2g, and must not be confused with 
that of the last two figures. For accuracy the second point 
should be as far from the orifice along the jet as possible. 

496. Orifice with Eounded Approach.* — Fig. 541 shows the 
general form and proportions of an orifice or mouth-piece in 
the use of which contraction does not 
take place beyond the edges, the inner N 
surface being one " of revolution," and 
so shaped that the liquid filaments are - 
parallel on passing the outer edge m ; ~Z-^^i~ " 
hence the pressure-head at m is = b - V 
(=34 ft. for water and 30 inches for 
mercury) in Bernoulli's Theorem, if 
efflux takes place into the air. We FlG - 541 - 

have also the sectional area F m = F — that of final edge of 
orifice, i.e., the coefficient of contraction, or O, = unity = 1.00, 
so that the discharge per time-unit has a volume 

Q = F m v m = Fv m . 

The tank being large, as in Fig. 540, Bernoulli's Theorem 
applied to m and n will give, as before, 





v m = V2gk 
as a theoretical result, while practically we write 



v m =<pV2gh, (1) 

and Q = F<pV~2g~h (2) 

As an average is found to differ little from 0.97 with this 
orifice, the same value as for an orifice in thin plate (§ 495). 

497. Problems in Efflux Solved by Applying Bernoulli's 
Theorem. — In the two preceding paragraphs the pressure- 
heads at sections m and n were each = p a -=- y = height of 

* Smooth conical nozzles for fire-steams give <p = .97 with h = press.- 
head -f veloc.-head at base of play-pipe ; see p. 833. 



664 



MECHANICS OF ENGINEERING. 



the liquid barometer = b ; but in the following problems this 
will not be the case necessarily. However, efflux is to take 
place through a simple orifice in the side of a large reservoir, 
whose upper surface (n) is very large, so that v n may be put 
= zero. 

Problem I. — Fig. 542. What is the velocity of efflux, v m , at 
the orifice m (i.e., at the contracted sec- 
tion, if it is an orifice in thin plate) 
of a jet of water from a steam-boiler, if 
the free surface at n is at a height = h 
above m, and the pressure of the steam 
over the water is jp n , the discharge tak- 
ing place into the air? 

Applying Bernoulli's Theorem to sec- 
tion m at the orifice [where the pres- 
sure-head is b and velocity-head v m * —■ %g (unknown)] and to 
section n at water-surface (where velocity-head = and pres- 
sure-head — j>n -4- y\ we have, taking m as a datum for poten- 
tial heads so that z m = and z n = A, 

2 

.-. v m =^2g[f-b + h\. . . . . (1) 




Fig. 542. 



Example. — Let the steam-gauge read 40 lbs. (and hence 
p n = 54.7 lbs. per sq. inch) and h = 2 ft. 4 in. ; required v m . 

Also if I^= % sq. in., in "thin plate," required the rate of 
discharge (volume). The temperature of saturated steam of 
the given tension must be 286° Fahr. [see foot of page 607]. 
The water is practically at the same temperature and hence 
of a heaviness, y, of 57.7 lbs. per cubic ft. (p. 518). 

From eq. (1) above, then, with ft. lb. and sec, noting that 
for this case b = [(14.7 -f- 144) -=- 57.7] feet, 



%, 



\/ 



2 X 32.2 



"54.7 X 144 14.7 X 144 



+ -1 

+ 12j 



81.1 ft. per sec. 



57.7 57.7 

theoretically 5 but practically 



PKOBLEMS OF EFFLUX — ORIFICES. 

v m = 0.97 X 81.1 = 78.6 ft. per sec. ; 

so that the discharge begins at the rate of 

Q = 0.64 Fv m = 0.64 X i . yir X 78.6 = 0.174 cub. ft. p. sec 

Problem II. — Fig. 543. With what velocity, v m , will water 
flow into the condenser C of a steam-engine where the tension 
of the vapor is p m , < one atmosphere, if ^^ ..*. 

h = the head of water, and the flow takes 
place through an orifice in thin plate ? 
Taking position m in the contracted section 
where the filaments are parallel, and the 
pressure therefore equal to that of the sur- § 
rounding vapor, viz.,p m , and position n in 
the (wide) free surface of the water in the 
tank, where (at surface) the pressure is one 






. .'/ AJR*. „* 



^=fe==- 



^Tt=^ 



Fig. 543. 



atmosphere [and 



En_r_ 



= h — 34 ft.] and velocity practically 



zero ; we have, applying Bernoulli's Theorem to n and m 9 tak- 
ing m as a datum level for potential heads (so that z n = h and 

3 m = 0), 



%9 y 



+ b + h, 



v m =^Zg[h + b-^ . . 



and Q = F m v m , 

as theoretical results. But practically we must write 

«U=0.97a/5 

and § = F m v m = 



h + h -y]> 



OFv r 



(1) 

® 

(3) 
(4) 



in which F — area of orifice in thin plate, and C = coefficient 
of contraction = about 0.62 approximately [see § 495]. 



666 



MECHANICS OF ENGINEERING. 



Example. — If in the condenser there is a "vacuum" of 27-§ 
inches (meaning that the tension of the vapor would support 
2£ inches of mercury, in a barometer), so that 

Pm = [f$ X 14.7] lbs. per sq. inch, and h = 12 feet, 

while the orifice is \ inch in diameter ; we have, using the ft., 
lb., and sec, 



<v m = 0.97a A X 32.2 fl2 + 34 



^ X 14.7 X 144 



62.5 



I 



= 51.1 ft. per sec. 
(We might also have written, for brevity, 



Pr 



m : 30] X 34 = 2.833, 



since the pressure-head for one atmos. = 34 feet, for water. 
Hence, for a circular orifice in thin plate, we have the volume 
discharged per unit of time, 

Q = CFv = 0.62 X j(t^X 51.1 = 0.0431 cub. ft. per sec. 



W 



497a. Efflux through an Orifice in Terms of the Internal and 
External Pressures. — Fig. 544. Let efflux take place through 
a small orifice from the plane side of a large tank, in which at 
the level of the orifice the hydrostatic pressure was =p' be- 
fore the opening of the orifice, that of the medium surround- 
ing the jet being =p". When a steady flow 
is established, after opening the orifice, the 
pressure in the water on a level with the ori- 
fice will not be materially changed, except in 
(o) EE5E L J:? the immediate neighborhood of the orifice [see 
ly Ss^c ^—Sa § 495] ; hence, applying Bernoulli's Theorem 
to m in the jet, where the filaments are parallel, 
and a point n, in the body of the liquid and 
at the same level as m, and where the particles 



.zzrftc: 



P- 



Fig. 544. 



are practically at rest [i.e., v n = 0] (hence not too near the 



FORCE-PUMP. 



667 



orifice), we shall have, cancelling out the potential heads which 
are equal, 

2g~ t y 2g^ y' 



v m = 0.97^9^ 



-P 



y 



(i) 



*m. 



(In Fig. 544 p' would be equal to p a -f- hy.) Eq. (1) is con- 
veniently applied to the jet produced by 
a force-pump, supposing, for simplicity, 
the orifice to be in the head of the purnp- 
cylinder, as shown in Fig. 545. Let the - 
thrust (force) exerted along the piston- 
rod be = P, and the area of the piston 
be = F' . Then the intensity of internal 
pressure produced in the chamber AB 
(when the piston moves uniformly) is 

,_P+F% 



Fig. 545. 



while the external pressure in the air around the jet is simply 
p a (one atmos.). 



°V^ 



(i)' 



(N.B. Of course, at points near the orifice the internal 
pressure is < p'\ read § 495.) 

Example. — Let the force, or thrust, i 3 , [due ta steam-pres- 
sure on a piston not shown in figure,] be 2000 lbs., and the 
diameter of pump-cylinder be d — 9 inches, the liquid being 
salt water (so that y = 64 lbs. per cubic foot). 

Then 

F' = i*(A) 2 = 0.442 sq. ft., 



and [ft., lb., sec] 



668 MECHANICS OF ENGINEEKITO. 

v m = 0.97^/2 x 32.2 X o^J^ = 65.4 ft. per se*. 

If the orifice is well rounded, with a diameter of one inch, 
the volume discharged per second is 

Q = F m v m = Fv m = |(^) 2 X 65.4 = 0.353 cub. ft. per sec. 

To maintain steadily this rate of discharge, the piston must 
move at the rate [veloc. = v'~\ of 






<o' = Q -s- F' = .353 -v 



|(~y = 0.800 ft. per sec, 



and the force P must exert a power (§ 130) of 

L = Pv f = 2000 X 0.800 = 1600 ft. lbs. per sec. 
= about 3 horse-power (or 3 H. P.). 

If the water must be forced from the cylinder through a 
pipe or hose before passing out of a nozzle into the air, the 
velocity of efflux will be smaller, on account of "fluid fric- 
tion" in the hose, for the same P; such a problem will be 
treated later [§ 513]. Of course, in a pum ping-engine, by the 
use of several pump-cylinders, and of air-chambers, a practically 
steady flow is kept up, notwithstanding the fact that the mo- 
tion of each piston is not uniform, and must be reversed at the 
end of each stroke. 

498. Influence of Density on the Velocity of Efflux in the Last 
Problem. — From the equation 



-=V 2 ^ 



of the preceding paragraph, where p" is the external pressure 
around the jet, and p' the internal pressure at the same level 
as the orifice but well back of it, where the liquid is sensibly 



RELATION OF DENSITY TO VELOCITY OF EFFLUX. 



at rest, we notice that for the same difference of pressure 
[jp'~~ P >r \ th e velocity of efflux is inversely proportional to th^ 
square root of the heaviness of the liquid. Hence, for the 
same {p r — p"\ mercury would flow out of the orifice with a 
velocity only 0.272 of that of water ; for 




Again, assuming that the equation holds good for the flow of 
gases (as it does approximately when j/ does not greatly exceed 
p"\ e.g., by 6 or 8 per cent), the velocity of efflux of atmospheric 
air, when at a heaviness of 0.807 lbs. per cub. foot, would be 



/ 62.5 

V .0807 



= V775.3 = 27.8 



times as great as for water, with the same p' — p'\ (See 
§ 548, etc.) 



499. Efflux under Water. Simple Orifice.— Fig. 546. Let h x 
and A 2 be the depths of the (small) ori- 
fice below the levels of the " head " and 
" tail " waters respectively. Then, using 
the formula of § 457, we have for the 
pressure at n (at same level as m, the 
jet) 

/ = & + %, 



n-AIR.- ._ 



:.¥. -. 



and for the external pressure, around 
the jet at m, 

whence, theoretically, 



Fig. 546. 



v m = ^2gZ 



—p" _ 



V2g{K-h t ) = V2gh, . (1) 



where h = difference of level between the surfaces of the two 
bodies of water. 



670 



MECHANICS OF ENGINEEKING. 



Practically, 



v m =<p Y2gh; 



(2) 



but the value of cp for efflux under water is somewhat uncer- 
tain ; as also that of O, the coefficient of contraction. Weis- 
hach says that pi, = 0(7, is -^ part less than for efflux into the 
air ; others, that there is no difference (Trautwine). See also 
p. 389 of vol. 6, Jour, of Engin. Associations, where it is 
stated that with a circular mouth-piece of 0.37 in. diam., and 
of " nearly the form of the vena contracta" jj, was found to be 
.952 for discharge into the air, and .945 for submerged dis- 
charge. 

500. Efflux from a Small Orifice in a Vessel in Motion. 

Case I. When the motion is a vertical translation and uni- 
formly accelerated. — Fig. 547. Suppose the vessel to move up- 
ward with a constant acceleration p. 
(See § 49a.) Taking m and n as in the 
two preceding paragraphs, we know that 
p m =z p" = external pressure = one at- 

^ = h). As to the 
Y 
internal pressure at n (same level as m, 
but well back of orihce), p n , this is not 
equal to- (b -f- h)y, because of the acceler- 
ated motion, but we may determine it by considering free the 
vertical column or prism On of liquid, of cross-section — dF, 
the vertical forces acting on which are p a dF, downward at 0, 
p n dF upward at n, and its weight, downward, hdFy. All 
other pressures are horizontal. For a vertical upward acceler- 
ation =p, the algebraic sum of the vertical components of all 
the forces must = mass X vert. acceL, 




mos. = p a (and 



Fig. 547. 



i.e., 
whence 



dF{p n -p a - hy) = JL— .p; 

if 



(1) 



Putting p n and p a equal to the p' and p" , respectively, of 
the equation, we have 



ORIFICE IN MOTION". 



671 



^VM^T*] 



of § 497, 



v m = y2{g+j>)h. 



(2) 



It must be remembered that v m is the velocity of the jet rel- 
atively to the orifice, which is itself in motion with a variable 
velocity. The absolute velocity w m of the particles of the jet 
is found by the construction in § 83, being represented graph- 
ically by the diagonal of a parallelogram one of whose sides is 
v m , and the other the velocity c with which the orifice itself is 
moving at the instant, as part of the vessel. The jet may 
make any angle with the side of the vessel. 

On account of the flow the internal pressures of the water 
against the vessel are no longer balanced horizontally, and the 
latter will swing out of the vertical unless properly constrained. 

If p = g = ace. of gravity, v m = V 2 V 2gh. If p is nega- 
tive and = g, v m = ; i.e., there is no flow, but both the vessel 
and its contents fall freely, without mutual action. 

Case II. When the liquid and the vessel both have a uni- 
form rotary motion about a vertical axis with an angular veloc- 
ity = go (§ 110). Orifice small, so that we may consider the 
liquid inside (except near the orifice) to 
be in relative equilibrium. Suppose the 
jet horizontal at m, Fig. 548, and the 
radial distance of the orifice from the 
axis to be = x. The external pressure 
p m — p a , and the internal [see § 428, 

eqs. (3) and (4)] is 



Pn=J?a+(h +z)r = Pa+hr 



2g 




Fig. 548. 



hence the velocity of the jet, relatively to the orifice, is (from 
§ 497, since p n and p m correspond to the p f and p' r of that 
article), 



v„ 



J^g (£=_£=> = V2gh + (Gov)*, 



672 MECHANICS OF ENGINEEKING. 



i.e., v m = V2gh + io*-, (3). 

in which w, = gdx, = the (constant) linear velocity of the ori- 
fice in its circular path. The absolute velocity w m of the par- 
ticles in the jet close to the orifice is the diagonal formed on 
w and v m (§ 83). Hence by properly placing the orifice in the 
casing, w m may be made small or large, and thus the kinetic 
energy carried away in the effluent water be regulated, within 
certain limits. Equation (3) will be utilized subsequently in 
the theory of Barker's Mill. 

Example. — Let the casing make 100 revol. per min. (whence 
go = [2tt100 -t- 60] radians per sec), h — 12 feet, and x = 2 
ft. ; then (ft., lb., sec.) 



v m = aA X 32.2 X 12 + ( 27r1 ^ X 2 ) 2 = 34.8 ft. per sec. 

(while, if the casing is not revolving, v m = V%gh = only 27.8 
ft. per sec). 

If the jet is now directed horizontally and backward, and 
also tangentially to the circular path of the centre of the orifice, 
its absolute velocity (i.e., relatively to the earth) is 

w m = v m — oox — 34.8 — 20.9 = 13.9 ft. per sec, 

and is also horizontal and backwards. If the volume of flow 
is Q = 0.25 cub. feet per sec, the kinetic energy carried away 
with the water per second (§ 133) is 

= i Mw m *=Qr.^ = iX 62 - 5 M = 46.8 
^2 32.2 2 

ft. lbs. per second = 0.085 horse-power. 

501. Theoretical Efflux tlirough Rectangular Orifices of Con- 
siderable Vertical Depth, in i Vertical Plate. — If the orifice is 
so large vertically that the velocities of the different filaments 
in a vertical plane of the stream are theoretically different, hav- 
ing different " heads of water," we proceed as follows, taking 
into account, also, the velocity of approach, c, or mean velocity 



RECTANGULAR ORIFICES. 673 

(if any appreciable), of the water in the channel approaching 
the orifice. 

Fig. 549 gives a section of the side of the tank and orifice. 
Let b = width of the rectangle, the sills of the latter being 
horizontal, and a = A 2 — A, , its height. Disregarding con- 
traction for the present, the theoretical volume of discharge 
per unit of time is equal to the 
sum of the volumes like v m dF 
{= v m bdx), in which v m == the 
velocity of any filament, as m, 
in the jet, and bdx = cross-sec- 
tion of the small prism which 
passes through any horizontal 
strip of the area of orifice, in a 
unit of time, its altitude being 
v m . For each strip there is a fig 549. 

different x or " head of water," and hence a different velocity. 
Now the theoretical discharge (volume) per unit of time is 

v m dF\ 

x=h t 

Q = o£v m dx (ly 




i.e, 



But from Bernoulli's Theorem, if k == c* -t- 2g = the velocity- 
head at n, the surface of the channel of approach nO,b being 
the pressure-head of n, and x its potential head referred to m as 
datum (N.B. This b = 34 ft. for water, and must not be con- 
fused with the width b of orifice), we have [see § 492, eq. (7)] 

/. v m = \fYg Y'V+W; (2)' 

and since dx = d(x -f- k), 7c being a constant, we have, from (1)' 
and (2)', 

Theoret. Q = b \ / lfy t f(x~+ k)id(x + Jc\ 



674 MECHANICS OF ENGINEERING. 

or 

Theoret. Q = V>V^g [(A 2 + Jc)* - (A, + k)*\ . . (1) 

(b now denotes the width of orifice.) If c is small, the chan- 
nel of approach being large, we have 

Theoret Q = \b V^g \hf -hf) .... (2) 

(c being = Q -— area of section of nO). 

If h x = 0, i.e., if the orifice becomes a notch in the side, or 
an overfall [see Fig 550, which shows the contraction which 
actually occurs in all these cases], we have for an overfall 

Theoret Q = %b V^g[(h 2 + yfc)t - #]. .... (3) 

Note. — Both in (1) and (2) \ and A 2 are the vertical depths 
...... . . .... . of the respective sills of the orifice 

from the surface of the water 
three or four feet back of the plane 
of the orifice, where the surface is 
comparatively level. This must 
be specially attended to in deriv- 
FlG - 55 °- ing the actual discharge from the 

theoretical (see § 503). 

If Q were the unknown quantity in eqs. (1) and (3) it would 
be necessary to proceed by successive assumptions and ap- 
proximations, since Q is really involved in k ; for 

h = ^ and F,e=Q 

(where F is the sectional area of the channel of approach nC). 
With k — (or c very small, i.e., F very large), eq. (3) re- 
duces (for an overfall) to 




Theoret Q = %bh 2 V2gh„ .... (3J) 

or f as much as if all parts of the orifice had the same head of 
water = A 2 (as for instance if the orifice were in the horizontal 
bottom of a tank in which the water was A 2 deep, the orifice 
having a width = b and length = A a ). 



TRIANGULAR ORIFICE. 



675 



502. Theoretical Efflux through a Triangular Orifice in a Thin 
Vertical Plate or Wall. Base Horizontal. — Fig. 551. Let the 
channel of approach be so large that the velocity of approach 
may be neglected. h 1 and A 2 = depths of sill and vertex, 
which is downward. The analysis diifers from that of the 
preceding article only in having k = and the length u, of a 
horizontal strip of the orifice, variable ; b being the length of 
the base of the triangle. From similar triangles we have 



u 



K — x 



- r ; i.e., u = 



h 



{K-x). 



b K — h; 

b r h * 

.\ Theoret. Q = fv m dF '= fv r jtdx = ^—-J^vjh^ — x)dx ; 
and finally, substituting from eq. (2)' of § 501, with Jc = 0, 





Fig. 551. 



Fig. 553. 



Theoret. Q = |_^ jT(X - «)a&fo 



= Y 5 %^jVW-MA* + 3h l q. ... (4) 



For a triangular notch as in Fig. 552, this reduces to 

25^' • W 



=4M,«=A^yp;; 



i.e., !% of the volume that would be discharged per unit of 



676 



MECHANICS OF ENGINEERING. 



time if the triangular orifice witli base b and altitude A 2 were 
cut in the horizontal bottom of a tank under a head of A 2 . 
The measurements of h 2 and b are made with reference to the 
level surface back of the orifice (see figure) ; for the water- 
surface in the plane of the orifice is curved below the level 
surface in the tank. 

Prof. Thomson has found by experiment that with 
b = 2A 2 , the actual discharge = theoret. disch. X 0.595 ; and 
with b = 4A 2 , actual = theoret. disch. X 0.620. 

503. Actual Discharge through Sharp-edged Rectangular Ori- 
fices (sills horizontal) in the vertical side of a tank or reservoir. 
Case I. Complete and Perfect Contraction. — The actual 
volume of water discharged per unit of time is much less than 

the theoretical values derived in § 501, 
chiefly on account of contraction. By 
complete contraction we mean that no 
edge of the orifice is flush with the 
side or bottom of the reservoir ; and 
by perfect contraction, that the channel 
of approach, to whose surface the 
heads h x and h 2 are measured, is so 
large that the contraction is practically 
the same if the channel were of infi- 

FlG " 553 ' nite extent sideways and downward 

from the orifice. 

For this case (A x not zero) it is found most convenient to 

use the following practical formula (b = width) : 




Actual Q = jA abA/2g\ K + ^ > 



(6) 



in which (see Fig. 553) a = the height of orifice, h x = the ver- 
tical depth of the upper edge of the orifice below the level of 
the reservoir surface, measured a few feet back of the plane of 
the orifice, and ju is a coefficient of efflux (an abstract number), 
dependent on experiment. 

With // = 0.62 approximate results (within 3 or 4 per cent) 
may be obtained from eq. (6) with openings not more than 



RECTANGULAR OEIFICES. 



677 



18 inches, or less than 1 inch, high ; and not less than 1 inch 



wide; with heads [h x -f- - J from 1 ft. to 20 or 30 feet. 



Example. — What is the actual discharge (volume) per min- 
ute through the orifice in Fig. 553, 14 inches wide and 1 
foot high, the upper sill being 8 ft. 6 in. below the surface of 
still water % Use eq. (6) with the ft., lb., and sec. as units, and 
fi 9 = 0.62. 

Solution : 
Q = 0.62 X 1 X H X V2 X 32.2[8J+i] = 17.41cub.ft.per.sec. 

while the flow of weight is 

G= Qy = 17.41 X 62.5 = 1088 lbs. per second. 

Poncelet and Lesbros 1 Experiments. — For comparatively ac- 
curate results, values of yw taken from the following table 
(computed from the careful experiments of Poncelet and Les- 
bros) may be used for the sizes there given, and, where prac- 
ticable, for other sizes by interpolation. To use the table, the 
values of h x , a, and b must be reduced to metres, which can be 
done by the reduction-table below; but in substituting in eq. 
(6), if the metre-kilogram-second system of units be used g 
must be put = 9.81 metres per. square second (see § 51), and Q 
will be obtained in cubic metres per second. 

Since pi is an abstract number, once obtained as indicated 
above, it does not necessitate any particular system of units in 
making substitutions in eq. (6). The ft., lb., and sec. will be 
used in subsequent examples. 

TABLE FOR REDUCING FEET AND INCHES TO METRES. 



1 foot 


= 0.30479 metre. 


1] 


nch 


= 0.0253 metre. 


2 feet 


= 0.60959 


u 


2 inches 


= 0.0507 


a 


3 " 


= 0.91438 


a 


3 


u 


= 0.0761 


n 


4 " 


= 1.21918 metres. 


4 


a 


= 0.1015 


a 


5 " 


= 1.52397 


u 


5 


it 


= 0.1268 


a 


6 " 


= 1.82877 


u 


6 


a 


= 0.1522 


a 


7 " 


= 2.13856 


a 


7 


u 


= 0.1776 


a 


8 " 


= 2.43836 


a 


8 


u 


= 0.2030 


a 


9 " 


= 2.74315 


a 


9 


a 


= 0.2283 


a 


10 " 


= 3.04794 


u 


10 


u 


= 0.2536 


u 








11 


a 


= 0.2790 


u 



678 MECHANICS OF ENGINEERING. 

TABLE, FROM PONCELET AND LESBROS. 

Values of juo, for Eq (6), for Rectangular Orifices in Thin Plate 

(Complete and perfect contraction.) 



Value of hi t 

Fig. 553 (in 

metres). 


b = .20™- 


b = .20 m - 


b = .20 m - 


6 = .20™- 


6 = .20™- 


b = .20 m - 


b = .60n>- 


b = .60*- 


a = .20 111 - 


a = .10™- 


a = .0^- 


a = ,03 m - 


a = .02 m - 


a = .0l m - 


a = ^O™- 


a = 02 m - 




Mo 


Mo 


Mo 


Mo 


Mo 


Mo 


Mo 


Mo 


0.005 












0.705 






.010 






0.607 


0.630 


0.660 


.701 




0.644 


.015 




0.593 


.612 


.632 


.660 


.697 




.644 


.020 


0.572 


.596 


.615 


.634 


.659 


.694 




.643 


.030 


.578 


.600 


.620 


.638 


.659 


.688 


0.593 


.642 


.040 


.582 


.603 


.623 


.640 


.658 


.683 


.595 


.642 


.050 


.585 


.605 


.625 


.640 


.658 


.679 


.597 


.641 


.060 


.587 


.607 


.627 


.640 


.657 


.676 


.599 


.641 


.070 


.588 


.609 


.628 


.639 


.656 


.673 


.600 


.640 


.080 


.589 


.610 


.629 


.638 


.656 


.670 


.601 


.640 


.090 


.591 


.610 


.629 


.637 


.655 


.668 


.601 


.639 


.100 


.592 


.611 


.630 


.637 


.654 


.666 


.602 


.639 


.120 


.593 


.612 


.630 


.636 


.653 


.663 


.603 


.638 


.140 


.595 


.613 


.630 


.635 


.651 


.660 


.603 


.637 


i .160 


.596 


.614 


.631 


.634 


.650 


.658 


.604 


.637 


.180 


.597 


.615 


.630 


.634 


.649 


.657 


.605 


.636 


.200 


.598 


.615 


.630 


.633 


.648 


.655 


.605 


.635 


.250 


.599 


.616 


.630 


.632 


.646 


.653 


.606 


.634 


.300 


.600 


.616 


.629 


.632 


.644 


.650 


.607 


.633 


.400 


.602 


.617 


.628 


.631 


.642 


.647 


.607 


.631 


.500 


.603 


.617 


.628 


.630 


.640 


.644 


.607 


.630 


.600 


.604 


.617 


.627 


.630 


.638 


.642 


.607 


.629 


.700 


.604 


.616 


.627 


.629 


.637 


.640 


.607 


.628 


.800 


.605 


.616 


.627 


.629 


.636 


.637 


.606 


.628 


.900 


.605 


.615 


.626 


.628 


.634 


.635 


.606 


.627 


1.000 


.605 


.615 


.626 


.628 


.633 


.632 


.605 


.626 


1.100 


.604 


.614 


.625 


.627 


.631 


.629 


.604 


.626 


1.200 


.604 


.614 


.624 


.626 


.628 


.626 


.604 


.625 


1.300 


.603 


.613 


.622 


.624 


.625 


.622 


.603 


.624 


1.400 


.603 


.612 


.621 


.622 


.622 


.618 


.603 


.624 


1.500 


.602 


.611 


.620 


.620 


.619 


.615 


.602 


.623 


1.600 


.602 


.611 


.618 


.618 


.617 


.613 


.602 


.623 


1.700 


.602 


.610 


.617 


.616 


.615 


.612 


.602 


.622 


1.800 


.601 


.609 


.615 


.615 


.614 


.612 


.602 


.621 


1.900 


.601 


.608 


.614 


.613 


.612 


.611 


.602 


.621 


2.000 


.601 


.607 


.613 


.612 


.612 


.611 


.602 


.620 


3.000 


.601 


.603 


.606 


.608 


.610 


.609 


.601 


.615 



Example. — With h, = 4 in. [ = 0.10 met.] , a = 8 in. 
[=0.20 met], 5 = 1 ft. 8 in. [=0.51 met], required the 
(actual) volume discharged per second. See Fig. 553. 



KECTANGTJLAK ORIFICES. 



679 



From the foregoing table, 
for A, = 0.10 m -, I = 0.60 m - and a = 0.20 m -, we find ^ = .602 



" h, = 0.10 ra -, b ±= 0.20 m - 



a 



0.20 1 



diff. 



/*, = .592 
= .010 



Hence, by interpolation, 

for h x = 0.10 m - b = 0.51 m -, and a = 0.20 m -, we have 



M Q = 0.602 



A [0.602 



0.592] = 0.600. 



Hence [ft., lb., sec], remembering that /* is an abstract num- 
ber, from eq. (6), 



Q = 0.600 X A X -H- ^2 X 32.2(^ + 1- A) = 4.36 

cub. ft. per second. 

Case II. Incomplete Contraction. — This name is given to 
the cases, like those shown in Fig. 554, where one or more 
sides of the orifice have an interior border flush with the sides 
or bottom of the (square-cornered) tank. 

Not only is the general direction of the stream altered, but 
the discharge is greater, on account of the larger size of the 
contracted section, since contraction is prevented on those sides 
which have a border. It is assumed that the contraction which 
does occur (on the other edges) is perfect ; i.e., the cross-sec- 
tion of the tank is large compared with the orifice. According 
to the experiments of Bidone and 
Weisbach with Poncelet's ori- 
fices (i.e., orifices in thin plate 
mentioned in the preceding table), 
the actual volume discharged per 
unit of time is 



Q = jxdb 



\M+l) 



<?) 




Fig. 554. 



(differing from eq. (6) only in 
the coefficient of efflux //), in which the abstract number ju is 
found thus : Determine a coefficient of efflux jn as if eq. (6) 
were to be used in Case I ; i.e., as if contraction were complete 
and perfect ; then write 



680 MECHANICS OF ENGINEERING. 

/< = /<o[l + 0.155 w], (7)' 

where n = the ratio of the length of periphery of the orifice 
with a border to the whole periphery. 
E.g., if the lower sill, only, has a border, 

while if the lower sill and both sides have a border, 
n = (2a + l>) + l2{a + by]. 

Example.— If h x = 8 ft. (= 2.43 m -), 3 = 2 ft. (= 0.60 m -), 
a = 4 in. (= 0.10 m -), and one side is even with the side of 
the tank, and the lower sill even with the bottom, required the 
volume discharged per second. (Sharp-edged orifice, in ver- 
tical plane, etc.) 

Here for complete and perfect contraction we have, from 
Poncelet's tables (Case I), /i = 0.608. Now n = i; hence, 
from eq. (7/, 

fi = 0.608 [1 + 0.155 X i]= 0.6551 ; 

hence, eq. (7), 

Q = 0.655 X 2 X A V2X 32.2(8+*. A) 

= 10.23 cub. ft. per sec. 

Case III. Imperfect Contraction. — If there is a submerged 

j_ ____ channel of approach, symmetrically 

placed as regards the orifice, and of 

_-l.-=~.-^^ c A, an area (cross-section), = G. not 

:zEEF-^ D^|A_ _i_ much larger than that, = F, of the 

r^. ^ ^li^i?^ orifice (see Fig. 555), the contraction 

"^JE^^M, ^ 1 is less than in Case I, and is called 

EEE^sfy imperfect contraction. Upon his 

" J l experiments with Poncelet's orifices, 

PrG# 555 - with imperfect contraction, Weisbach 

bases the following formula for the discharge (volume) per 

unit of time, viz., 



= /««A»/W(a,+ !) ..... (8) 



KECTANGULAR OKIFICES. 



681 



(see Fig. 553 fci notation), with the understanding that the co< 
efficient 

M = rtl+fi), (8)' 

where /* is the coefficient obtained from the tables of Case I 
(as if the contraction were perfect and complete), and /? an ab- 
stract number depending on the ratio F : G = ra, as follows : 



/? = 0.0760 [9 m - 1.00], 



(8) / 



Table A. 



To shorten computation Weisbach gives the following table 
for/?: 

Example.— Let \ = 4' %" (= 1.46 
met.), the dimensions of the orifice 
being — 

width = h = 8 in. (= 0.20 m ); 
height = a = 5 in. (= 0.126 m ) ; 

while the channel of approach {CD, 
Fig. 555) is one foot square. From 
Case I, we have, for the given di- 
mensions and head, 

ja q = 0.610 ; 



m. 


£. 


m. 


J8. 


.05 


.009 


.55 


.178 


.10 


.019 


.60 


.208 


.15 


.030 


.65 


.241 


.20 


.042 


.70 


.278 


.25 


.056 


.75 


.319 


.30 


.071 


.80 


.365 


.35 


.088 


.85 


.416 


.40 


107 


.90 


.473 


.45 


.128 


.95 


.537 


.50 


.152 


1.00 


.608 



G 144 sq. in. 



We find [Table A] 



P == 0.062 



and hence v = Mo (1-062), from eq. (8)'. Therefore, from eq. 

(8), with ft, lb., and sec, 



Q = 0.610 X 1.062 XA-AV2X 32.2 X 5 

= 3.22 cub. ft. per sec. 

Case IV. Head measured in Moving Water, — See Fig. 
556. If the head A, , of the upper sill, cannot be measured to 
the level of still water, but must be taken to the surface of a 
channel of approach, where the velocity of approach is quite 



682 



MECHANICS OF ENGINEEKING. 



appreciable, not only is the contraction imperfect, but 
strictly we should use eq. (1) of § 501, in 
which the velocity of approach is considered. 
Let F = area of orifice, and G that of the 
cross-section of the channel of approach; 
then the velocity of approach is c = Q -i- G, 
and k (of above eq.) == & -f- 2g = Q* -r- 2^6r 2 ; 
but Q itself being unknown, a substitution of 
h in terms of Q in eq. (1), § 501, leads to an 
equation of high degree with respect to Q. 

Practically, therefore, it is better to write 




Q = M^2g{h 1 + a -y 



(9) 



and determine M by experiment for different values of the 
ratio F-r- G. Accordingly, Weisbach found, for Poncelet's 
orifices, that if /* is the coefficient for complete and perfect 
contraction from Case I, we have 

I* = /<.(!+/»') (»)' 

fi' being an abstract number, and being thus related toF-±- G, 



/3' = 0.641 



(9)" 



A, was measured to the surface one metre back of the plane of 

the orifice, and F : G did not exceed 0.50. 

Weisbach gives the following table computed from eq. (9)" : 
Table B. Example. — A rectangular water-trough 4 ft. 
wide is dammed up with a vertical board in 
which is a rectangular orifice, as in Fig. 556, of 
width 5 = 2 ft. (= 0.60 met.), and height a = 6 
in. (= 0.15 met.) ; and when the water-level be- 
hind the board has ceased rising (i.e., when the 
flow has become steady), we find that h 1 = 2 ft., 
and the depth behind in the trough to be 3 ft. 
Eequired Q. 

Since F: G = l sq. ft. -f- 12 sq. ft. = .0833, 

we find (Table B) /?' — 0.005 ; and jj being = 0.612 from Pon 

celet's tables, Case I, we have finally, from eq. (9), 



F+G. 


0'. 


0.05 


.002 


.10 


.006 


.15 


.014 


.20 


.026 


.25 


.040 


.30 


.058 


,35 


.079 


,40 


.103 


.45 


.130 


.50 


.160 



DISCHAKGE OF OVEKFALL-WEIRS. 



683 



Q = 0.612(1.005) 2 xi V2 X 32.2 X 2.25 
= 7.41 cub. f-t. per second. 

5U4. Actual Discharge of Sharp-edged Overfalls (Overfall- 
weirs; or Rectangular Notches in a Thin Vertical Plate). 

Case I. Complete and Perfect Contraction {the normal 
case), Fig. 557 ; i.e., no edge is flush 
with the side or bottom of the 4f§s^V^ 

reservoir, whose sectional area is ^r^' 

very large compared with that, M 2 , -^^ 
of the notch. By depth, h 2 , of the 
notch, we are to understand the 
depth of the sill Mow the surface 
a few feet hack of the notch where 
it is level. In the plane of the 
notch the vertical thickness of the stream is only from f to -£$ 
of A a . Putting, therefore, the velocity of approach = zero, 
and hence Tc = 0, in eq. (3) of § 501, we have for the 




Fig. 557. 



Actual Q = jJt<%bK vtyK , 



(10) 



Table C. 



(b = width of notch,) where // is a coefficient of efflux to be 
determined by experiment. 

Experiments with overfalls do not agree as well as might be 
desired. Those of Poncelet and Lesbros gave the results in 
Table C. 

Example 1. — With 

\ = 1 ft. 4 in. (= .405 m -), 
o = 2 ft. (= 0.60 m -), 

we have, from Table C, /i = .586, 
and (ft., lb., sec.) 



.-. #=.586xfX2X# |/2X32.2X| 
= 9.54 cub. ft. per sec. 

Example 2. — What width, 5, 
must be given to a rectangular notch, for which h 2 = 10 in. 
(= 0.25 m -), that the discharge may be Q = 6 cub. feet per sec. 'I 



For 6 = 0.20°*. 


For & = 


= 0.60 m . 


metres. 


metres. 




h-t fi 6 


h<i 


fJ-o 


.01 .636 


.06 


.618 


.02 .620 


.08 


.613 


.03 .618 


.10 


.609 


.04 .610 


.12 


.605 


.06 .601 


.15 


.600 


.08 .595 


.20 


.592 


.10 .592 


.30 


.586 


.15 .589 


.40 


.586 


.20 .585 


.50 


.586 


.22 .577 


.60 


.585 


For approx. i 


esults Mo 


= .60 



684 



MECHANICS OF ENGINEEKING. 



Since b is unknown, we cannot use the table immediately, 
but take /j = .600 for a first approximation ; whence, eq. (10), 
(ft., lb., sec.,) 



b = 



6 



0.6XIXH1/2X 32.2 X If 



= 2.46 ft. 



Then, since this width does not much exceed 0.60 metre, 
we may take, in Table C, for A a = 0.25 met., /* = .589 ; 



b = 



6 



.589 X I X n v% X 32.2 X & 



2.50 ft. 



Case II. Incomplete Contraction * i.e. , both ends are flush 
with the sides of the tank, these being ~] to the plane of the 
notch. According to Weisbach, we may write 

Q = %Mbh 2 Vtyh„ (11) 

in which pi = 1.041/i , /* being obtained from Table C for the 
normal case, i.e., Case I. The section of channel of approach 
is large compared with that of the notch ; if not, see Case IV. 
Case III. Imperfect Contraction; i.e., the velocity of ap- 
proach is appreciable ; the sectional area G 
of the channel of approach not being much 
larger than that, F, = bh 2 = area of notch. 
Fig. 558. b = width, and h, 2 = depth of 
notch (see Case I). Here, instead of using 
a formula involving 




Jc = c* + 2g = [Q+G]> + 2g 



Fig. 558. 



(see eq. (3), § 501), it is more convenient to put 
as before, with 



Q=iMbh a V2gh„ (12) 



M = Mo (l + 0), (12)' 

in which /* is for the normal case [Case I] ; and /?, according 



OVERFALL- WEIRS . 



685 



to Weisbach's experiments, may be obtained from the empiri- 
cal formula 



/3 = 1.718 



©• 



(12)" 



Table D. 



[Table D is computed from (12) // .] 

(The contraction is complete in this case ; i.e., the ends are 
not flush with the sides of the tank.) 

Example. — If the water in the channel of ap- 
proach has a vertical transverse section of G = 9 
sq. feet, while the notch is 2 feet wide (i.e., 
b = 2') and 1 foot deep (A 2 = V) (to level of 
surface of water 3 or 4 ft. back of notch), we 
have, from Table C, with b = .60 met. and 
A, = 0.30 met., 

^ = 0.586; 
while from Table D, with F : G = 0.222 (or -§), 

£ = .005; 
hence (ft.-lb.-sec. system of units), from eq. (12), 



F 




G 


/3. 


0.05 


.000 


.10 


.000 


.15 


.001 


.20 


.003 


.25 


.007 


.30 


.014 


.35 


.026 


.40 


.044 


.45 


.070 


.50 


.107 



Q = f X 0.586 X 1.005 X 2 X 1 X V64.4 X 1.0 
= 6.30 cub. ft. per second. 

Case IV. Fig. 559. Imperfect and incomplete contrac- 
tion together / both end-contractions being "suppressed" (by 
making the ends flush with the sides of the reservoir, these 
sides being vertical and ~] to the plane of the notch), and the 
channel of approach not being very deep, i.e., having a sec- 
tional area G but little larger than that, F, of notch. F= bk t 
as before. 

Again we write 

Q = Mh 9 Vfy£„ (13) 

with }x computed from 

J M = ^(l + /S), (13)' 



yw being obtained from Table C ; while 



686 



MECHANICS OF ENGINEERING. 



/3 = 0.041 + 0.3693 [?S , .... (13)" 

an empirical formula based by Weisbach on his own experi- 
ments.* To save computation, j3 may be found from Table E, 
founded on eq. (13)". 

Table E. 



F 
G~ 

= 


.00 


.05 


.10 


.15 


.20 


.25 


.30 


.35 


.40 
.100 


.45 
.116 


.50 


.041 


.042 


.045 


.049 


.056 


.064 


.074 


.086 


.133 




Example.— Fig. 559. With 
5 = 2 ft. (=0.60 met.) 



and 



A, = 1 ft. (= 0.30 met.), 



we have, from Table C, M = 0.586. 
But, the ends being flush with the 
sides of the reservoir or channel, 

Fig. 559. and Q being = q gq ft ( gee fi gure ) ) 

which is not excessively large compared with i' 7 = &A a = 2 sq. 
ft., we have, from Table E, with F : G = 0.333, 

/? = .081; 

and hence [eq. (13) and (13)'], ju being .586 as in last example, 



Q = | X 0.586 X (1 + .081) X 2 X 1 X VteA X 1.0 

== 6.78 cub. ft. per sec. 

505. Francis' Formula for Overfalls (i.e., rectangular notches). 
— From extensive experiments at Lowell, Mass., in 1851, with 
rectangular overfall-weirs, Mr. J. B. Francis deduced the fol- 
lowing formula for the volume, Q, of flow per second over 
such weirs 10 feet in width, and with A 2 varying from 0.6 to 
1.6 feet (from sill of notch to level surface of water a few feet 
back) : 



* Weisbach's results in this case differ considerably from those of Bazin- 

see p. 688. 



OVERFALL-WEIRS. 687 



Q=iX0.622h t (b-f ls nh t )V2gfi t > • • ( 14 ) 

in which b = width. 

This provides for incomplete contraction, as well as for com- 
plete and perfect contraction, by making 

n = 2 for perfect and complete contraction (Fig. 557) ; 
n = 1 when one end only is flush with side of channel ; 
n = when both ends are flush with sides of channel. 

The contraction was considered complete and perfect when 
the channel of approach was made as wide as practicable, 
= 13.96 feet, the depth being about 5 feet. 

Mr. Francis also experimented with submerged or " drowned" 
weirs in 1883 ; such a weir being one in which the sill is be- 
low the level of the tail-water (i.e., of receiving channel). 

506. Fteley and Stearns's Experiments at Boston, Mass., in 1877 
and 1880. — These may be found in the Transactions of the 
American Society of Civil Engineers, vol. xn, and gave rise 
to formulse differing slightly from those of Mr. Francis in 
some particulars. In the case of suppressed end-contractions, 
like that in Fig. 559, they propose formulae as follows : 

When depth of notch is not large, 

Q (in cub. ft. per sec.) = 3.31 bh% + 0.007 1 . (15) 

(b and h 2 both in feet), 

" A 2 , the depth on the weir, should be measured from the sur- 
face of the water above the curvature of the sheet." 

" Air should have free access to the space under the sheet." 
The crest must be horizontal. The formula does not apply to 
depths on the weir less than 0.07 feet. 

When the depth of notch is quite large, a correction must 
be made for velocity of approach, o, thus : 

Q (in cub. ft. per sec.) = 3.31 b\h, -f 1.5 |^T+ 0.007 1 (16) 

(b and h % both in feet). 



MECHANICS OF ENGINEERING. 

The channel should be of uniform rectangular section for 
about 20 ft. or more from the weir, to make this correction 
properly. If G = the cross-section, in sq. ft., of the channel 
of approach, c is found approximately by dividing an approxi- 
mate value of Q by G ; and so on for closer results. 

The weir may be of any length, &, from 5 to 19 feet. 

506a. Recent Experiments on Overfall-weirs in France. — In 

the Annates des Ponts et Chaussees for October 1888 is an 
account of extensive and careful experiments conducted in 
1886 and 1887 by M. Bazin on the flow over sharp-edged 
overfall- weirs with end-contractions suppressed ; i.e., like that 
shown in Fig. 559. The widths of the weirs ranged from 
0.50 to 2.00 metres, and the depths on the weirs (h 2 ) from 
0.05 to 0.60 metre. With p indicating the height of the sill 
of the weir from the bottom of the channel of approach, M. 
Bazin, as a practical result of the experiments, recommends 
the following formula as giving a reasonably accurate value 
for the volume of discharge per unit of time : 

$ = f|i'[l+0.6S(^J]M.VP; f . . (17) 

where the coefficient ji' has a value 

^'=°- 6075 +^l)-- ; < 18 > 

Eq. (17) is homogeneous, i.e., admits of any system of units. 

Provision was made in these experiments for the free en- 
trance of air under the sheet (a point of great importance), 
while the walls of the channel of approach were continued 
down-stream, beyond the plane of the weir, to prevent any 
lateral expansion of the sheet. The value of j> ranged from 
0.20 to 2.00 metres. 

Herr Ritter von Wex in his " Hydrodynamilc " (Leipsic, 
1888) derives formulae for weirs, in the establishing of which 
some rather peculiar views in the Mechanics of Fluids are 
advanced. 



OYEKFALLS. 



689 



Formulae and tables for discharge through orifices or over 
weirs of some forms not given here may be found in the 
works of Weisbach, Rankine, and Trau twine. 

Mr. Hamilton Smith, a noted American hydraulic engineer, 
lias recently published " Hydraulics," a valuable compilation 
and resume of the most trustworthy experiments in all fields 
of hydraulics (New York, 1886 : John Wiley & Sons). 



507. Efflux through Short Cylindrical Tubes. — When efflux 
takes place through a short cylindrical tube, or " short pipe," 
at least 2 \ times as long as wide, 
inserted at right angles in the 
plane side of a large reservoir, 
the inner corners not rounded 
(see Fig. 560), the jet issues 
from the tube in parallel fila- 
ments and with a sectional area ? 
F m , equal to that, F, of interior 
of tube. 

To attain this result, however, FlG - 56 °- 

the tube must be full of water before the outer end is un- 
stopped, and must not be oily; nor must the head, A, be 
greater than about 40 ft. for efflux into the air. Since at m 
the filaments are parallel and the pressure-head therefore equal 
to b (== 34 ft. of water, nearly), = that of surrounding medium, 
= head due to one atmosphere in this instance ; an application 
of Bernoulli's Theorem [eq. (7), § 492] to positions m and n 
would give (precisely as in §§ 454 and 455) 



... < -;ajr';: v .; .-?. 




^eM0^ 


i 
i 


— fv ^ 


' i.'-l'? 


"~-'N \ 


i -j " 


sSjJ! 




i- . 





v m = veloc. at m = V%gh 

as a theoretical result ; but experiment shows that the actual 
value of v m in this case is 



<D m =<t> o V2gh = 0.S15tf2gh, . 



(1) 



0.815 being an average value for O , the coefficient of 'velocity ', for 
ordinary purposes. It increases slightly as the head decreases, 



MECHANICS OF ENGINEERING. 

and is evidently much less than the value 0.97 for an orifice in 
a thin plate, § 495, or for a rounded mouth-piece as in § 496. 

But as the sectional area of the stream where the filaments 
are parallel, at m, where v m = 0.815 V2gh, is also equal to that, 
F, of the tube, the coefficient of efflux, /* , in the formula 



Q = Mo FV2gh, 

is equal to O ; i.e., .there is no contraction, or the coefficient 
of contraction, C, in this case == 1.00. 

Hence, for the volume of discharge per unit of time, we 
have practically 

Q = 4> FV2gh = 0M5FV2ghl ... (2) 

The discharge is therefore about -J greater than through an 
orifice of the same diameter in a thin plate under the same 
head [compare eq. (3), § 495] ; for although at m the velocity 
is less in the present case, the sectional area of the stream is 
greater, there being no contraction. 

This difference in velocity is due principally to the fact that 
the entrance of the tube has square edges, so that the stream 

contracts (at m', Fig. 561) to a 
section smaller than that of the 
tube, and then re-expands to the 
full section, F, of tube. The 
eddying and accompanying in- 
ternal friction caused by this re- 
expansion (or "sudden enlarge- 
"-'■'' ment" of the stream) is the prin- 
FlG * 561, cipal resistance which diminishes 

the velocity. It is noticeable, also, in 
this case that the jet is not limpid and 
clear, as from thin plate, but troubled 
and only translucent (like ground- 
glass). The internal pressure in the 
stream at m' is found to be less than 
one atmosphere, i.e. less than that at m 9 
as shown experimentally by the suck- Fig. 562. 

ing in of air when a small aperture is made in the tube op- 





INCLINED SHORT PIPES. 



691 



posite m r . If the tube itself were so formed internally as to 
lit this contracted vein, as in Fig. 562, the eddying would be 
diminished and the velocity at m increased, and hence the 
volume Q of efflux increased in the same proportion. (See 
§ 509a.) 

If the tube is less than 2J- times as long as wide, or if the 
interior is not wet by the water (as when greasy), or if the head 
is over 40 or 50 ft. (about), the efflux takes 
place as if the tube were not there, Fig. 563, 
and we have ^ 




v m = 0.97 V2gh, as in § 495. 

Example. — The discharge through a short 
pipe 3 inches in diameter, like that in Fig. 560, 
is 30 cub. ft. per minute, under a head of 
2' 6", reservoir large. Required the coefficient of efflux 
ju , = O , in this case. For variety use the inch-pound-min- 
ute system of units, in which g = 32.2 X 12 X 3600 (see Note, 
§ 51). // , being an abstract number, will be the same numer- 
ically in any system of units. 

From eq. (2), 



0o 


= M = 

= 0.802 


Q - 

FV2gh 


30 X 1728 




71 






j X 3 2 V2 X 32.2 X 12 X 60 2 


X30 



508. Inclined Short Tubes (Cylindrical).— Fig. 564 
short tube is inclined at some angle 
a < 90° to the interior plane of the 
reservoir wall, the efflux is smaller than 
when the angle is 90°, as in § 507. 

We still use the form of equation 



If the 



Q = M FV2 g h = <pFV2gh; . (3) 

but from "Weisbach's experiments fi 
should be taken from the following table: 




Fig. 564. 



692 MECHANICS OF ENGINEEKING. 

TABLE F, COEFFICIENT OF EFFLUX (Inclined Tube). 



For a = 90° 
take ju = 0= .815 



80° 
799 



70° 

,782 



60° 
764 



50° 

,747 



40° 
731 



30° 
.719 



Example. — With h = 12 ft., d = diam. of tube = 4 ins., 
and a = 46°, we have for the volume discharged per sec. (ft., 
lb., and sec.) 

Q = [0.731 + -JL (.016)] ? fcj V64.4X12=1.W cub.ft.per sec. 

The tube must be at least 3 times as long as wide, to be 
filled. 

509. Conical Diverging, and Converging, Short Tubes. — "With 
conical convergent tubes, as at A, Fig. 565, with inner edges 
not rounded, D'Aubuisson and Castel found by experiment 
values of the coefficient of velocity, 0, and of that of efflux, /*, 
[from which the coefficient of contraction, O = jx -j- 0, may be 



air ■.••■ ;.:.'.•.*.•;.' 




Fig. 565. 

computed,] for tubes 1.55 centimeters wide at the narrow end, 
and 4.0 centimeters long, under a head of h = 3 metres, and 
with different angles of convergence. By angle of converg- 
ence is meant the angle between the sides CE and DB. Fig. 
565. In the following table will be found some values of pi 
and (p founded on these experiments, for use in the formulae 



v m = V2gh and Q = jxF V2gh ; 
in which ^denotes the area of the outlet orifice EB. 



CONICAL SHORT PIPES. 



693 



Table G (Conical Converging Tubes). 



Angle of ) = 3 o 1(y 
convergence ) 

ju = .895 
<p= .894 



.932 



10° 20' 



,951 



13° 30' 

.946 
.963 



19° 30' 

.924 
.970 



30° 

.895 
.975 



49° 

,847 



Evidently jjl is a maximum for 13J°. 

With a conically divergent tube as at MN, having the in- 
ternal diameter MO = .025 metre, the internal diam. NP 
= .032 metre, and the angle between MN said PO_= 4° 50', 
Weisbach found that in the equation Q = piF V2gh (where 
F — area of outlet section NP) pi should be = 0.553; the 
great loss of velocity as compared with V2gh being due to the 
eddying in the re-expansion from the contracted section at M 
(corners not rounded), as occurs also in Fig. 549. The jet was 
much troubled and pulsated violently. 

When the angle of divergence is too great, or the head h too 
large, or if the tube is not wet by the water, efflux with the 
tube filled cannot be maintained, the flow then taking place as 
in Fig. 563. 

Venturi and Eytelwein experimented with a conically di- 
vergent tube (called now " Ven- 
turis tube"), with rounded en- 
trance to conform to the shape 
of the contracted vein, as in 
Fig. 566, having a diameter of 
one inch at m / (narrowest part), 
where the sectional area = F' 
= 0.7854 sq. in., and of 1.80 
inches at m (outlet), where area = F\ the length being 8 ins., 
and the angle of convergence 5° 9'. 

With Q = /*F¥2gh they found a = 0.483. 

Hence 2J times as much water was discharged as would have 
flowed out under the same head through an orifice in thin 
plate with area = F' = the smallest section of the divergent 
tube, and 1.9 times as much as through a short pipe of sec- 
tion == F '. A similar calculation shows that the velocity at 
m' must have been v m > = 1.55 V2gh, and hence that the pres- 
sure at m' was much less than one atmosphere. 




Fig. 566. 



694 



MECHANICS OF ENGINEERING. 



Mr. J. B. Francis also experimented with Yenturi's tube 
(see " Lowell Hydraulic Experiments"). See also p. 389 of 
vol. 6 of the Journal of Engineering Societies, for experi- 
ments with diverging short tubes discharging under water. 
The highest coefficient (/*) obtained by Mr. Francis was 0.782. 

509a. New Forms of the Venturi Tube.— The statement made 
in § 507, in connection with Fig. 562, was based on purely 
theoretic grounds, bat has recently (Dec. 1888) been com- 
pletely verified by experiments* conducted in the hydraulic 
laboratory of the College of Civil Engineering at Cornell 
University. Three short tubes of circular section, each 3 in. 
in length and 1 in. in internal diameter at both ends, were ex- 
perimented with, under heads of 2 ft. and 4 ft. Call them A, 
B, and C. A was an ordinary straight tube as in Fig. 561 ; 
the longitudinal section of B was like that in Fig. 562, the 
narrowest diameter being 0.80 in. [see § 495 ; (0.8) 2 = 0.64] ; 
while C was somewhat like that in Fig. 566, being formed 
like B up to the narrowest part (diameter 0.80 in.), and then 
made conically divergent to the discharging end. The results 
of the experiments are given in the following table : 



Name of 
Tube. 



A 
A 

B 
B 

C 

C 



Head. 



Number 
of Experi- 
ments. 



ft = 2 ft. 
ft = 4 ft. 

ft = 2 ft. 

ft = 4 ft. 

ft = 2 ft. 
ft = 4 ft. 



Range of Values of y.. 



From 0.804 to 0.823 
" 0.819 to 0.823 

" 0.875 to 0.886 
" 0.881 to 0.902 

" 0.890 to 0.919 
" 0.902 to 0.923 



Average 
Values of /*. 



0.814 
0.821 

0.882 
0.892 

0.901 
0.914 



The fact that B discharges more than A is very noticeable, 
while the superiority of C to B, though evident, is not nearly 
so great as that of B to A, showing that in order to increase 
the discharge of an (originally) straight tube (by encroaching 
on the passage-way) it is of more importance to fill up with 
solid substance the space around the contracted vein than to 
make the transition from the narrow section to the discharg- 
ing end very gradual. 

* See Journal of the Franklin Inst., for April, 1889. 



"fluid friction." 



695 



510. "Fluid Friction."— By experimenting with the flow of 
water in glass pipes inserted in the side of a tank, Prof Bey 
nolds of England has found that the flow goes on in parallel 
filaments for only a few feet from the entrance of the tube 
and that then the liquid particles begin to intermingle and 
cross each other's paths in the most intricate manner To 
render this phenomenon visible, he injected a fine stream of 
colored liquid at the inlet of the pipe and observed its further 
motion, and found that the greater the velocity the nearer to 
the inlet was the point where the breaking up of the parallel 
ism of flow began. The hypothesis of laminated flow is 
nevertheless, the simplest theoretical basis for establishing 
practical formula, and the resistance offered by pipes to the 
flow of liquids in them will therefore be attributed to the fric 
tion of the edges of the lamina against the inner surface of 
the pipe. 

The amount of this resistance (often called skin-friction) 
for a given extent of rubbing surface is by experiment found— 

1. To be independent of the pressure between the liquid and 
the solid ; 

2. To vary nearly with the square of the relative velocity; 

3. To vary directly with the amount of rubling surface ■ ' 

. 4. To vary directly with the heaviness [y, § 409] of 'the 

Hence for a given velocity v, a given rubbing surface of 
area = &, and a liquid of heaviness y, we may write 

Amount of friction (force) =fSy — , .... m 

in which/ is an abstract number called the coefficient of fluid 
friction, to be determined by experiment. For a oiven liquid 
given character (roughness) of surface, and small range of 
velocities it is approximately constant. The object of intro- 
ducing the 2g is not only because ~ is a familiar and useful 

function of «, but that v* ~- 2g is a height, or distance, and there- 
fore the product of S (an area) by v 2 - 2g is a volume, and this 
volume multiplied by y gives the weight of an ideal prism of 



MECHANICS OF ENGINEERING. 



the liquid ; hence S — y is & force and f must be an abstract 

number and therefore the same in all systems of units, in any 
given case or experiment. 

In his experiments at Torquay, England, the late Mr. Froude 
found the following values for f, the liquid being salt water, 
while the rigid surfaces were the two sides of a thin straight 
wooden board -^ of an inch thick and 19 inches high, coated 
or prepared in various ways, and drawn edgewise through the 
water at a constant velocity, the total resistance being measured 
by a dynamometer. 

511. Mr. Froude' s Results. — (Condensed.) [The velocity was 
the same = 10 ft. per sec. in each of the following cases. For 
other velocities the resistance was found to vary nearly as the 
square of the velocity, the index of the power varying from 
1.8 to 2.16.] 

Table H. 



Character of Surface. 


Value of / [from eq. (1), § 510]. 




2 ft. long. 


When the 
8 ft. long. 


board was 
20 ft. long. 


50 ft. long. 


Varnish ./ = 

Paraffine " 


0.0041 
.0038 
.0030 
.0087 
.0081 
.0090 
.0110 


0.0032 
.0031 
0028 
.0063 
.0058 
.0062 
.0071 


0.0028 
.0027 
.0026 
.0053 
.0048 
.0053 
.0059 


0.0025 


Tinfoil 


.0025 


Calico 


.0047 


Fine Sand 


.0040 


Medium Sand 


.0049 


Coarse Sand 








N.B. These numbers multiplied 
lbs. per sq. foot of area of surface 
heaviness of sea water, 64 lbs. per 
eq. (1) of the preceding paragraph. 


by 100 also give the mean frictional resistance in 
i in each case (v = 10' per sec.),, considering the 
cubic foot, to cancel the"2cr = 64.4 ft. per sq. sec. of 



For use in formulge bearing on flow in pipes, f is best deter, 
mined directly by experiments of that very nature, the results 
of which will be given as soon as the proper formulge have been 
established. 

512. Bernoulli's Theorem for Steady Flow, with Friction. — [The 
student will now re-read the first part of § 492, as far as eq. 
(1).] Considering free any lamina of fluid, Fig. 567, (according 
to the subdivision of the stream agreed upon in § 492 referred 



BERNOULLI'S THEOREM WITH FRICTION. 



697 



to,) the frictions on the edges are the only additional forces as 
compared with the system in Fig. 
534. Let w denote the length 
of the wetted perimeter of the 
base of this lamina (in case of a 
pipe running full, as we here 
postulate, the wetted perimeter 
is of course the whole perimeter, 
but in the case of an open chan" 
nel or canal, w is only a portion 
of the whole perimeter of the Y?ydi 

cross-section). Then, since the fio. 567. 

area of rubbing surface at the edge is 8= wds', the total fric- 
tion for the lamina is [by eq. (1), § 510] =fwy (V -~ 2g)ds'. 
Hence from vdv = (tan. accel.) X ds, and from (tan. accel.) — 
[^"(tang. com pons, of acting forces)] -f- (mass of lamina), we 
have 




vdv = 



Fp — F( p -\- dp) -f- Fyds' cos <p —fwy — ds f 

*9 .ds.Ja) 



Fyds' ~- g 

As in § 492, so here, considering the simultaneous advance of 
all the laminae lying between any two sections m and n during 
the small time dt, putting ds' = ds, and ds' cos <p = — dz (see 
Fig. 568), we have, for any one lamina, 



w 



- vdv A — dp-\-dz — — f-^=f 
a v J F 



ds. 



g ■ y - ' * JF fy 

"Now conceive an infinite number of equations to be 

like eq. (1), one for each la- _ 

mina between n and m, for the 

same dt, viz., a dt of such 

length that each lamina at the 

end of dt will occupy the 

same position, and acquire the 

same values of v, z, and p, 

that the lamina next in front 

had at the beginning of the 



• a) 

formed 




S\J* 



— '?> 



Fig. 568. 



dt (this is the characteristic of a steady flow). Adding up 



MECHANICS OF ENGINEEKING. 



the corresponding terms of all these equations, we have (re 
membering that for a liquid y is the same in all laminae), 

\Sfa + \J?* +/*=-£• /"J** ; • ( 2 ) 

i.e., after transposition and writing R for F-t-w, for brevity, 



2? 



+ & -K 









(3) 



This is Bernoulli's Theorem for steady flow of a liquid in 
a pipe of slightly varying sectional area F, and internal perim- 
eter w, taking into account no resistances or friction, except 
the " skin-friction," or u fluid-friction," of the liquid and sides 
of the pipe. 

Resistances due to the internal friction of eddying occasioned 
by sudden enlargements of the cross-section of the pipe, by 
elbows, sharp curves, valve-gates, etc., wiJl be mentioned later. 
The negative term on the right in (3) is of course a height or 
head (one dimension of length), as all the other terms are such, 
and since it is the amount by which the sum of the three heads 
(viz., velocity-head, pressure-head, and potential head) at m, 
the down-stream position, lacks of being equal to the sum of 
the corresponding heads at n, the up-stream position or section, 
we may call it the "Loss of Head" due to skin-friction between 
n and m; also called friction-head, or resistance-head, or 
height of resistance. 

The quantity R = F -f- w = sectional-area -=- wetted-pe 
rimeter, is an imaginary distance or length called the Hydrau- 
lic Mean Radius, or Hydraulic Mean Depth, or simply 
hydraulic radius of the section. For a circular pipe of diam- 
eter = d, 

R = \nd? -r- nd = \d ; 
while for a pipe of rectangular section, 

ao 



R = 



2{a+b)' 



FEICTION IN" PIPES. 



699 



513. Problems involving Friction-heads; and Examples of 
Bernoulli's Theorem with Friction. 

Problem I. — Let the portion of pipe between n and m be 
level, and of uniform cir- 






^ 




Fig. 569. 



cular section and diameter 
= d. The jet at m dis- 
charges into the air, and 
has the same sectional area, 
F= %7td%2LS the pipe; then 
the pressure-head at m is 

l^ = b = 34 feet (for 

r 

water), and the velocity- 
head at m is = that at n, since v m = v n . The height of the 
water column in the open piezometer at n is noted, and = y n 

(so that the pressure-head at n is^ = y n -{-b); while the 

r 

length of pipe from n to m is = I. 

Knowing I, d, y n , and having measured the volume Q, of 
flow, per unit of time, it is required to find the form of the 
friction-head and the value off From 



F m v m = Q, or '\ntfv m = Q, . 



(1) 



v m becomes known. Also, v m is known to be = v n , and the 
velocity at each ds is v = v m , since F (sectional area) is con- 
stant along the pipe, and Fv — Q. The hydraulic radius is 



R = \d 



(2) 



the same for all the ds's between n and m. 

Substituting in eq. (3) of § 512, with the horizontal axis of 
the pipe as a datum for potential heads, we have 

+ b + = ^ + y n + h + 0-^.^jris; . (3) 

i.e., since /ds = l = length of pipe from n to m, the friction- 
head for a pipe of length = I, and uniform circular section 
of diameter = d, reduces to the form 






700 



MECHANICS OF ENGINEEEING. 



Friction-head = 4/ 



A ?L 

d '2g 



• (4) 



where w = velocity of water in the pipe, being in this case 
also = v m and = v n . Hence this friction-head varies directly 
as the length and as the square of the velocity ', and inversely 
as the diameter / also directly as the coefficient/. 

From (3), then, we derive (for this particular problem) 

7 2 

Piezometer-height at n = y n = 4/- . — ; . . (5) 

t/ 

i.e., the open piezometer-height at n is equal to the loss of head 
(all of which is friction-head here) sustained between n and the 
mouth of the pipe. (Pipe horizontal.) 

Example. — Bequired the value of f, knowing that d = 3 in., 
y n (by observation) = 10.4 ft., and Q = 0.1960 cub. ft. per 
sec, while I = 400 ft. (n to m). From eq. (1) we find, in ft.- 
lb.-sec. system, the velocity in the pipe to be 



noJ 



4 X 0.1960 



7T- 1 
71 16" 



= 4.0 ft. per sec; 



then, using eq. (5), we determine/* to be 

f - %gy n d __ 2 X 32.2 X}X 10.4 
/ ~ "W 4 X 400 X 4* ~ °-° 065 - 

Problem II. Hydraulic Accumulator. — Fig. 570. Let the 
area F n of the piston on the left be quite large compared with 




Fig. 570. 



that of the pipes and nozzle. The cylinder contains a friction 






FRICTION-HEAD m PIPES. 



701 



less weighted piston, producing (so long as its downward slow 
mot.cn is uniform) a fluid pressure on its lower face of an 
intensity Fn = IG + F^ + F n per unit area ( Pa = one 
atmos.). y ^ 

Hence the pressure-head at n is 

T~F-? +h > ( 6 ) 

where G = load on piston. 

The jet has a section at m = F m = that of the small straight 
nozzle (no contraction). The junctions of the pipes with efch 
other, and with the cylinder and nozzle, are all smoothly 
rounded ; hence the only losses of head in steady flow between 
n and m are the friction-heads in the two long pipes nerfect 
•ng that in the short nozzle. These friction-heaKfl £ of 
the form m eq. (4), and will involve the velocities v, and v 
respectively in these pipes {supposed running full). v and v 
may be unknown at the outset, as here 

to ?ndTn Dg ^ a ^ a " d r e f i0DS aDd hei ^ hts ' ™ are required 

voW n < £ Vm ° f * e M ^S into the «fr, and the 
volume of flow, Q, per unit of time, assuming /to be known 

and to be the same in both pipes (not strictly true) 

Let the lengths and diameters be denoted as in Fig 570 

their sectional areas F x and F„ the unknown velocities in them 

From the equation of continuity [eq. (3), § 490], we have 
«, = ^ and »,= ^. . . . ( 7) 

ea T ?vt*ll m \Z e aP f^ ? ern ° nlli ' s Theore *> (™& Miction), 
eq. (3), 512, taking the down-stream position m in the jet 
c ose to the nozzle and the up-stream position n just under the 



702 MECHANICS OF ENGINEERING:. 

Apparently (8) contains three unknown quantities, v m , v x % 
and v t ; but from eqs. (7) v k and v 2 can be expressed in terms 
of v m , whence [see also eq. (6)] 

or, finally, 

and hence we have also 

Q = F m v m (11) 

Example. — If we replace the force G of this problem by 
the thrust P exerted along the pump-piston of a steam fire- 
engine, we may treat the foregoing as a close approximation 
to the practical problem of such an apparatus, the pipes being 
consecutive straight lengths of hose, in which (for the probable 
values of v x and v 2 ) we may take/ = .0075 (see " Fire-streams," 
by G-eo. Ellis, Springfield, Mass.).* (Strictly, f varies somewhat 
with the velocity; see § 517.) Let P= 12000 lbs., and the 
piston-area at n = F n = 72 sq. in. = -J sq. ft. Also, let h = 20 
ft., and the dimensions of the hose be as follows : 

d x = 3 in., d % = 2 in., d n (of nozzle) = 1 in. ; 

l x = 400 ft., l % = 500 ft. 

"With the foot-pound-second system of units, we now have 
[eq. (10)] 




2 X 32.2 



■ J2000J 

. T £X62.5. 



i+**.»H7l'+f(jy 



See § 578. 



FRICTION-HEAD IN PIPKS. 



703 



= . Ax 32.2 X 404 . 



1 + 0.59 -f 5.62 
i.e., v m = 60.0 ft. per sec. If this jet were directed vertically 
upward it should theoretically attain a height = ?• = nearly 

40 or1'5 b f t ^ r68iStanCe ° f the * W0UW r6duCe ^ t0 about 
We have farther, from eq. (1), 

Q = F m v m = | (I) ° x eo.o = 3.27 cub. ft. per sec. 
§ 4 I 97l here W6re n ° reSiStaDCe iU the h ° Se W Sh0uld have ' from 

"*" = y 2ff [£? + k ] = ^2>^3E2X404 = 161.3 ft. per sec. 

513a. Influence of Changes of Temperature—Although Poi- 
semlle and Hagen found that with glass tubes of very sma 1 
d.ameter the flow of water was increased threefold S 
temperature from 0° to 45° Cent., it is unlikely that with com 
mon p lpes the rate of flow is appreciably affecid by the ordi- 
nary fluctuations of temperature ; at any rate, experiments of 
sufficient precision are wanting, as regards suchT nfl„ nee 
See Mr. Hamilton Smith's "Hydraulics" r, Ifi I T ' 
naw "r'l 1 a„^„„ k • .- . iJ -£ ulduucs ) P- lb, where he 

says Changes by var.at.on in T (temperature) will probablv 
only be appreciable with small orifices, or with very low tads 
for orifices or weirs." ' 

814. loss of Head in Orifices and Short Pipes—So W as th» 
steady flow between two localities „ and m Les pkee in a p pe 

Wg^ abrupt enlargement or diminution of sectTon uor 
harp curv be ds> or e]bowg; the logs Qf ect op jor 

solely to the surface action (or « skin-f riction") beLeeu wate r 

and p.pe; but the introduction of auy of the aWmentToS 

eatures occasmns eddying aud iuternal disturbai.ee anc l7Z 

*» (and consequent heat): thereby causing furtheT'd Watil 



704 MECHANICS OF ENGINEEKING. 

from Bernoulli's' Theorem; i.e., additional losses of head, 01 
heights of resistance. 

From the analogy of the form of a friction-head in a long 
pipe [eq. (4), § 513], we may assume that any of the above 
heights of resistance is proportional to the square of the veloc- 
ity, and may therefore always be written in the form 

j Loss of Head due to any \ _ ~ v* , 

( cause except skin-friction ) ~~ ~ 2g 9 v ' 

in which v is the velocity of the water in the pipe at the sec- 
tion where the resistance occurs ; or if, on account of an 
abrupt enlargement of the stream-section, there is a correspond- 
ing diminution of velocity, then v is always to denote this 
diminished velocity (i.e., in the down-stream section). This 
velocity v is often an unknown at the outset. 

C, corresponding to the abstract factor 4f --m the height of 

Co 

resistance due to skin-friction [eq. (4), § 513], is an abstract 
number called the Coefficient of Resistance, to be determined 
experimentally ; or computed theoretically, where possible. 
Roughly speaking, it is independent of the velocity, for a given 
fitting, casing, pipe-joint, elbow, bend, valve-gate at a definite 
opening, etc., etc. 

515. Heights of Resistance (or Losses of Head) Occasioned by 
Short Cylindrical Tubes. — When dealing with short tubes dis- 
charging into the air, in § 507, deviations from Bernoulli's 
Theorem were made good by using a coefficient of velocity 0, 
dependent on experiment. This device answered every pur- 
pose for the simple circumstances of the case, as well as for 
simple orifices. But the great variety of possible designs of a 
compound pipe (with skin-friction, bends, sudden changes of 
cross-section, etc.) renders it almost impossible, in such a pipe, 
to provide for deviations from Bernoulli's Theorem by a single 
coefficient of velocity (velocity of jet, that is) for the pipe as a 
whole, since new experiments would be needed for each new 
design of pipe. Hence the great utility of the conception of 
" loss of head," one for each source of resistance. 



LOSS OF HEAD IN A SHORT PIPE. 705 

If a long pipe issues from the plane side of a reservoir and 
the corners of the junction are not rounded [see Fig. 571], we 

shall need an expression for 
the loss of head at the en- 
trance, E, as well as that 




[>v 



d '.9g 



FlG ' 571, due to the skin-friction in the 

pipe. But, whatever the velocity, v, in the pipe proves to 
be, influenced as it is both by the entrance loss of head and 
the skin-friction head (in applying Bernoulli's Theorem), the 

loss of head at E, viz., Ze 5—5 will be just the same as if efflux 

took place through enough of the pipe at E to constitute a 
" short pipe," discharging into the air, under some head h 
(different from h' of Fig. 571) sufficient to produce the same 
velocity v. But in that case we should have 

v 



v=<p V2gh, or Ji, = <f>*h. . . . . (1) 
2g 

(See §§ 507 and 508, being the " coefficient of velocity," and 
h the head, in the cases mentioned in those articles.) 

We therefore apply Bernoulli's Theorem to the cases of 
those articles (see Figs, 560 and 564) in order to determine the 
loss of head due to the short pipe, and obtain (with m as datum 
level for potential heads) 

gL+j + o-O + J + A-^!-. • • • (2) 

Now the v of eq. (2) is equal to the v m of the figures referred 
to, and C E is a coefficient of resistance for the short pipe, and 
we now desire its value. Substituting for 



2(7 L 2<7 J 



706 MECHANICS OF ENGINEERING. 

its value 2 A from eq. (1), we have 



<t> 



(3) 



Hence when a = 90° (i.e., the pipe is ~| to the inner reser- 
voir surface), we derive 



C - 1 1- 1 

** <P: (0.815)' 



1 = 0.505 ; . . [« = 90°;] . . (4) 



and similarly, for other values of a (taking <p from the table, 
§ 508), we compute the following values of Q E (corners not 

rounded) for use in the expression for " loss of head," Z E — : 



For a = 90° 


80° 
.565 


70° 
.635 


60° 
.713 


50° 
.794 


40° 30° 
.870 .987 



From eq. (4) we see that the loss of head at the entrance of 
the pipe, corners not rounded, with a = 90°, is about one half 
(,505) of the height due to the velocity v in that part of the 
pipe (v being the same all along the pipe if cylindrical). The 
value of v itself, Fig. 571, depends on all the features of the 
design from reservoir to nozzle. See § 518. 

If the corners at E are properly rounded, the entrance loss of 
head may practically be done away with; still, if v is quite 
small (as it may frequently be, from large losses of head 
farther down-stream), the saving thus secured, while helping 
to increase v slightly (and thus the saving itself), is insignifi 
cant. 

516. General Form of Bernoulli's Theorem, considering all 
Losses of Head, 

In view of preceding explanations and assumptions, we may 
write in a general and final form Bernoulli's Theorem for a 
steady flow from an up-stream position n to a down-stream 
position m, as follows : 

(all losses of head ) 
-\-z n — \ occurring between v . (Bj) 



ty 



\-2*+s, 






n and m 






LOSSES OF HEAD IN GENERAL. 707 

Each loss of head (or height of resistance) will be of the form 

2 7 a \ 

C ^— (except skin-friction head in long pipes, viz., 4/ — — J , 

the v in each case being the velocity, known or unknown, in 
that part of the pipe where the resistance occurs (and hence 
is not necessarily equal to v m or v n ). 

517. The Coefficient,/, for Friction of Water in Pipes. — See 
eq. (1), § 510. Experiments have been made by Weisbach, 
Eyteiwein, Darcy, Bossut, Prony, Dubuat, Fanning, and oth- 
ers, to determine f in cylindrical pipes of various materials 
(tin, glass, zinc, lead, brass, cast and wrought iron) of diameters 
from % inch up to 36 inches. In general, the following deduc- 
tions may be made from these experiments : 

1st. f decreases when the velocity increases ; e.g., in one 
case with the 

same pipe/" was = .0070 for v = 2' per sec, 
while/ was == .0056 for v = 20' per sec. 

2dly. f decreases slightly as the diameter increases (other 
things being equal); 

e.g., in one experiment/* was = .0069 in a 3-in. pipe, 

while for the same velocity f was = .0064 in a 6-in. pipe. 

3dly. The condition of the interior surface of the pipe 
affects the value of /, which is larger with increased roughness 
of pipe. 

Thus, Darcy found, with a foul iron pipe with d = 10 in. 
and veloc. = 3.67 ft. per sec, the value .0113 for/; whereas 
Fanning (see p. 238 of his " Water-supply Engineering"), with 
a cement-lined pipe and velocity of 3.74 ft. per sec. and d = 
20 inches, obtained/ = .0052. 

Weisbach, finding the first relation very prominent, pro- 
posed the formula 

.00429 
/ = 0.00359 + — ===== 
V v (in it. per sec.) 

when the velocities are great, while Darcy, taking into account 
both the 1st and 2d relations above, writes (see p. 585, Ran- 
kine's Applied Mechanics) 



708 



MECHANICS OF ENGINEERING. 



/=.0043 1 + — 



iam. in ft. J 



+ 



.001 



ft. per sec. 



[»+ 



18 X diam. in 



s] 



For practical purposes, Mr. J. T. Fanning has recommended, 
and arranged in an extensive table (pp. 242-246 of his book 
just mentioned), values of f for clean iron pipe, of diameters 
from \ inch to 96 inches, and for velocities of 0.1 ft. to 20 ft. 
per second. Of this the table opposite is an abridgment, in- 
serted with Mr. Fanning's permission, for use in solving nu- 
merical problems. 

In obtaining/ for slightly tuberculated and for foul pipes, 
the recommendations of Mr. Fanning seem to justify the fol- 
lowing rules : 



For slightly tuberculated pipes of diams. = \ ft. 

we should add 23$ 
and for foul pipes of same size 72$ 



lft.l 

34$ I 
60$ | 



2 ft, 4 ft. 

13$ 
25$ 



of they for clean pipes, to itself. For example, iff = .007 
for a certain f-ft. pipe when clean, with velocity = 0.64 ft. per 
sec, we have/= .007 X 1.72 = .01204 when it is foul. 

For first approximations a mean value off= .006 may be 
employed, since in some problems sufficient data may not be 
known in advance to enable us to take f from the table. 

Example. — Fig. 572. In the steady pumping of crude 
petroleum weighing y = 55 lbs. per cubic foot, through a six- 

,£ inch pipe 30 miles long, 
to a station 700 ft. higher 
than the pump, it is 
found that the pressure 
in the pump cylinder at 
n, necessary to keep up 
a velocity of 4.4 ft. per 
sec. in the pipe, is 1000 
lbs. per sq. inch. Eequired the coefficient f in the pipe. As 
all losses except the friction-head in the pipe are insignificant, 
the latter only will be considered. The velocity-head at n may 



? 




Fig. 572. 



TABLE OF VALUES OF /. 



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tH 


*-' 


C3 



710 MECHANICS OF ENGINEERING. 

be put = ; the jet at m being of the same size as the pipe, 
the velocity in the pipe is = v m , and therefore v m = 4.4 ft. per 
sec. Notice that m, the down-stream section, is at a higher 
level than n. 

From Bernoulli's Theorem, § 516, we have, with w as a 
datum level, 

Using the ft., lb., and sec, we have 

h = 700 ft., v m * + 2g = 0.30 ft., 
while 

5 = M^XJM = 38.47 ft., and & = ^^ = 2618 ft. 

Hence, in eq. (1), 

0.30 + 38.5 + 700 = 2618 - 4/. 80 X 528 ° . ^ . 
r ' - • i 64.4 

Solving fory, we have f = .00485 (whereas for water, with 
v = 4 .4 ft. per sec. and d = % ft., the table, p. 709, gives 
/= .00601. 

If the y °f- the petroleum had been 50 lbs. per cubic foot, 

instead of 55, we would have obtained £z = 2880 feet and f 
= .0056. 

518. Flow through a Long Straight Cylindrical Pipe, including 

both friction-head and entrance loss of head (corners not rounded); 

^ ^ reservoir large. Fig. 573. 

fe^-c"^ The jet issues directly 

i from the end of the pipe, 

_l £.'--*; in parallel filaments, into 

^^^ = ^ =l ~^\r~ ^-"^^""^ 8 the air, and therefore 
-iH^ has same section as pipe ; 

Fig. 573. — , , - ., . ' . 

hence, also, v m of the jet 
= v in the pipe (which is assumed to be running full), and is 



COEFFICIENT OF LIQUID FRICTION. 711 

V 9 

therefore the velocity to be used in the loss of head Q E — at 

if 

the entrance .#(§ 515). 

Taking m and n as in figure and applying Bernoulli's 
Theorem (§ 474), with m as datum level for the potential heads 
z m and z n , we have 

v 4+ b + o = o+b+h '"4'- 4f iw • (1) 

Three different problems may now be solved: 

First, required the head h to keep up a flow of given volume 

= Q per unit of time in a pipe of given length I and diameter 

= d. 
From the equation of continuity we have 

4:0 

.". veloc. of jet, which = veloc. in pipe, = v m = —~. . . (2) 

Having found v m = v, from (2), we obtain from (1) the re- 
quired h, thus : 

h =i\}+ z *+v l d\ ^ 

Now Q E = 0.505 if a = 90° (see § 515), while / may be 
taken from the table, § 517, for the given diameter and com- 
puted velocity [v m = v, found in (2)], if the pipe is clean ; if 
not clean, see end of § 517, for slightly tuberculated and for 
foul pipes. 

Secondly. Given the head h, and the length I and diameter 
d of pipe, required the velocity in the pipe, viz., v, = v m , that 
of jet ; also the volume delivered per unit of time, Q. Solv- 
ing eq. (1) for v m , we have 



/ 

\J 1 + ^+4/ 



/- — *V; ... (4) 



d 



712 MECHANICS OF ENGINEERING. 

whence Q becomes known, since 

Q = \nd\ m (5) 

[Note. — The first radical in (4) might for brevity be called 
a coefficient of velocity, 0, for this case. Since the jet has the 
same diameter as the pipe, this radical may also be called a 
coefficient of effimx.~\ 

Since in (4) f depends on the unknown v as well as on the 
known d, we must first puty = .006 for a first approximation 
for v m ; then take a corresponding value for f and substitute 
again ; and so on. 

Thirdly, knowing the length of pipe and the head A, we 
wish to find the proper diameter d for the pipe to deliver a 
given volume Q of water per unit of time. Now 

^>*-'*l&» w 

which substituted in (1) gives 

that is, 

W = (^-) 2 [(l + Q^ + 4/7]; 

As the radical contains d, we first assume a value for d, 
withy= .006, and substitute in (7). With the approximate 
value of d thus obtained, we substitute again with a new value 
for f based on an approximate v from eq. (6) (with d = its 
first approximation), and thus a still closer value for d is de- 
rived ; and so on. (Trautwine's Pocket-book contains a table 
of fifth roots and powers.) If I is quite large, we may put 
d = for a first approximation. In connection with these 
examples,' see last figure. 



LONG PIPES. 713 

Example 1. — What head h is necessary to deliver 120 cub* 
it. of water per minute through a clean straight iron pipe 140 
ft. long and 6 in. in diameter ? 

From eq. (2), with ft., lb., and sec, we have 

v = v m =[4:X W] *■ *(*)'= 10 - 18 ft - P er sec - 

Now for v = 10 ft. per sec. and d — \ ft., we find (in table, 
§ 517) /= .00549 ; and hence, from eq. (3), 



h = W™L [~1 + .505 + iX^0549><140- 

2 X 32.2 L ^ ^ i 



= 12.23 ft,, 



of which total head, as we may call it, 1.60 ft. is used in pro- 
ducing the velocity 10.18 ft. per sec. (i.e., v m * -h 2g = 1.60 ft.), 

while 0.808 ft. f = C^^-J is lost at the entrance ^(with a = 

90°), and 9.82 ft. (friction-head) is lost in skin-friction. 

Example 2. — [Data from Weisbach.] Required the de- 
livery, Q, through a straight clean iron pipe 48 ft. long and 
2 in. in diameter, with 5 ft. head (= A), v, =v m , being un- 
known, we first tdkef= .006 and obtain [eq. (4)] 



V ™= / 1 | 4x.006x48 ^ 2 >< 32 - 2 X 5 

» 6 

= 6.18 ft. per sec. 

From the table, § 517, for v = 6.2 ft. per sec. and d = 2 in. 9 
/= .00638, whence 



°-y 



I 1 

i . , A , , 4x .00638 X 48 ^ 2 x 32 ' 2 x 5 
1 -f- .505 -\ 



== 6.04 ft. per sec, 
which is sufficiently close. Then, for the volume per second, 

Q = j d 2 v m = J-7r(£) 2 6.04 = 0.1307 cub. ft. per sec. 



714 MECHANICS OF ENGINEERING. 

fWeisbach's results in this example are 
v m = 6.52 ft. per sec. 
and Q = 0.1420 cub. ft. per sec, 

but bis values f or/ are slightly different.] 

Example 3. — [Data from Weisbach.] What must be the 
diameter of a straight clean iron pipe 100 ft. in length, which 
is to deliver Q = \ of a cubic foot of water per second under 
5 ft. head (= A)? 

With/= .006 (approximately), we have from eq. (7), put- 
ting d = under the radical for a first trial (ft., lb., sec), 



, 5 /4 X .006 X 100 /f \ 2 , . A Q . ,. 
d = V 2X32.2X5 * (S = ab ° Ut °- 3 ° ft 5 

whence -y = — ^ = 7 ft. per sec 



For rf = 3.6 in. and v = 7 ft. per sec, we find/= .00601 ; 
whence, again, 

d= 6 / 1.505 X .30 + 4 X .00601 X 100 (4 X ?VL 0j3 24ft.. 

\/ ^i X o2i.& X 5 \ 7t I 

and the corresponding u = 6.06 ft. 

For this <# and w we find/= .00609, whence, finally, 

a _ s / 1.505 X .30 + 4 X .00609 X 100 (2\* . o a n 
d -\J 2 X 32.2 X 5 \n) = °- 326 fL 

[Weisbach's result is d = .318 ft.] 

519. Ch6zy's Formula. — If, in the problem of the preceding 
paragraph, the pipe is so long, and therefore I : d so great, 
that 4/7 -T- d in eq. (3) is very large compared with 1 -f- C* > 
we may neglect the latter term without appreciable error; 
whence eq. (3) reduces to 

7 2 

A = 4/--.||- . . (pipe very long ; Fig. 573), . . (8) 



ch£zy's foemula. 715 

which is known as Cheztfs Formula. For example, if I = 1000 
ft. and d = 2 in. = -§- ft., and/approx. = .006, we have 4/- = 

144, while 1 + C E for square corners = 1.505 only. 
If in (8) we substitute 



(8) reduces to 






h== -^'f#'ty • • • (^ery long pipe); .... (9) 

so that for a very long pipe, considering f as approximately 
constant, we may say that to deliver a volume = Q per unit 
of time through a pipe of given length = I, the necessary head, 
A, is inversely proportional to the fifth power of the diameter. 

And again, solving (9) for Q, we find that the volume con- 
veyed per unit of time is directly proportional to the fifth power 
of the square root of the diameter / directly proportional to 
the square root of the head ; and inversely proportional to the 
square root of the length. (Not true for short pipe ; see above 
example.) 

If we conceive of the insertion of a great nnmber of piezom- 
eters along the long straight pipe, of uniform section, now 
under consideration, the summits of the respective water 
columns maintained in them will lie in a straight line joining 
the discharging (into the air) end of the pipe with a point in 
the reservoir surface vertically over the inlet extremity (prac- 
tically so), and the " slope" of this line (called the Hydraulic 
Grade Line or Gradient), i.e., the tangent (or sine ; the angle 
is so small, generally) of the angle which it makes with the 

horizontal is = —, and may be denoted by s. Putting also 



yi = B, = the hydraulic radius of the section of the pipe, and 
v m = v = velocity in pipe, we may transform eq. (8) into 



v = \/% (&)* ; or, v = A(Bs)\ . . . (10) 




716 



MECHANICS OF ENGINEERING. 



which is the form by which Mr. Hamilton Smith (see § 506) 
interprets all the experiments quoted by him on long pipes. 
As to notation, however, he uses n for A, and r for jR. With 
the foot and second as units, the quantity A (not an abstract 
number) varies approximately between 60 and 140. For a 
given A we easily find the corresponding f from the relation 

f—-^. If the pipe discharges under water, h = the differ- 
ence of elevation of the two reservoirs. If the pipe is not 
horizontal, the use of the length of its horizontal projection 

instead of its actual length in the relation s = 



I 



occasions an 



error, but it is in most cases insignificant. 

Similarly, if a steady flow is going on in a long pipe of uni- 
form section, at the extremities of any portion of which we 
have measured the piezometer heights (or computed them 
from the readings of steam or pressure gauges), we may apply 
eq. (9), putting for h the difference of level of the piezometer 
summits, and for I the length of the pipe between them. 

520. Coefficient / in Fire-engine Hose.* — Mr. Geo. A. Ellis, 
C.E., in his little book on " Fire-streams," describing experi- 
ments made in Springfield, Mass., gives a graphic comparison 
(p. 45 of his book) of the friction-heads occurring in rubber 
hose, in leather hose, and in clean iron pipe, each of 2 J in. 
diameter, with various velocities; on which the following state- 
ments may be based : That for the given size of hose and 
pipe (d — %i in.) the coefficient f for the leather and rubber 
hose respectively may be obtained approximately by adding to 
f for clean iron pipe (and a given velocity) the per cent of 
itself shown in the accompanying table. 

Example. — For a clean iron pipe 
2£ in. diam., for a velocity = 10 ft. 
per sec, we have, from § 517, f ■=• 
.00593. Hence for a leather hose of 
the same diameter, we have, for v = 
10 ft. per sec, 

/= .00593 + .43 X .00593 = .00848. 

* For the most recent and exhaustive experiments in this direction see 
Mr. J. R. Freeman's "Hydraulics of Fire-streams" in the Transac. Am. 
Soc. C. E. for 1889. (Also p. 832 of this work.) 



Velocity 


Rubber 


Leather 


ft. per 


bose 2£ in. 


hose 2£ in . 


sec. 


diam. 


diam. 


3.0 


50$ 


300£ 


6.5 


20 


80 


10 


16 


43 


13 


12.5 


82 


16 


12 


30 



PRESSTJRE-ENEKGY. 



717 



521. Bernoulli's Theorem as an Expression of the Conservation 
of Energy for the Liquid Particles. — In any kind of flow with- 
out friction, steady or not, in rigid immovable vessels, the 
aggregate potential and kinetic energy of the whole mass of 
liquid concerned is necessarily a constant quantity (see §§ 148 
and 149), but individual particles (as the particles in the sink- 
ing free surface of water in a vessel which is rapidly being 
emptied) may be continually losing potential energy, i.e., 
reaching lower and iower levels, without any compensating in- 
crease of kinetic energy or of any other kind ; but in a steady 
flow without friction in rigid motionless vessels, we may state 
that the stock of energy of a given particle, or small collection 
of particles, is constant during the flow, provided we recognize 
a third kind of energy which may be called Pressure-energy, 
or capacity for doing work by virtue of internal fluid pressure ; 
as may be thus explained : 

In Fig. 574 let water, with a very slow motion and under a 
pressure p (due to the reservoir-head -|- atmosphere-head be- 
hind it), be admitted behind a pis- 
ton the space beyond which is 
vacuous. Let s == length of 
stroke, and F = the area of pis- 
ton. At the end of the stroke, 
by motion of proper valves, com- 
munication with the reservoir is 
cut off on the left of the piston 
and opened on the right, while the water in the cylinder now on 
the left of the piston is put in communication with the vacu- 
ous exhaust-chamber. As a consequence the internal pressure 
of this water falls to zero (height of cylinder small), and on 
the return stroke is simply conveyed out of the cylinder, 
neither helping nor hindering the motion. That is, in doing 
the work of one stroke, viz., 

W = force X distance = Fp X s — Fps, 

a volume of water V= Fs, weighing Fsy (lbs. or other unit), 
has been used, and, in passing through the motor, has experi- 
enced no appreciable change in velocity (motion slow), and 



l^<~ 



« 



*» 



& 



Fig. 574. 



718 MECHANICS OF ENGINEERING. 

therefore no change in kinetic energy, nor any change of level, 
and hence no change in potential energy, but it has given up 
all its pressure. (See § 409 for y.) 

Now W, the work obtained by the consumption of a weight 
= G = Vy of water, may be written 

W=Fps = Fsp= Vp = Vy^=G^. . . (1) 

Hence a weight of water = G is capable of doing the work 
G X — = G x head due to pressure p, i.e., = G x pressure- 

y 

head, in giving up all its pressure p ; or otherwise, while still 
having a pressure p, a weight G of water possesses an amount 
of energy, which we may call pressure-energy, of an amount 

= G' — , where y = the heaviness (§ 7) of water, and - = a 

y y 

height, or head, measuring the pressure p ; i.e., it equals the 
pressure-head. 

We may now state Bernoulli's Theorem without friction in 
a new form, as follows : Multiplying each term of eq. (7), 
§ 451, by Qy, the weight of water flowing per second (or other 
time-unit) in the steady flow, we have 

Qy % + Qy 7 s + QyZm = Qr W + Qr f + Qy2n ' (2) 

v 3 1 Qy 
But Qy £- = — -^- v m * = i X mass flowing per time-unit X 

2g 2 g 

square of the velocity = the kinetic energy inherent in the 
volume Q of water on passing the section m, due to the veloc- 
ity at m. Also, Qy— = the pressure-energy of the volume 

Q at m, due to the pressure at m ; while Qyz m = the potential 
energy of the volume Q at m due to its height z m above the 
arbitrary datum plane. Corresponding statements may be 
made for the terms on the right-hand side of (2) referring to 
the other section, n, of the pipe. Hence (2) may be thus read : 
The aggregate amount of energy (of the three kinds mentioned) 
resident in the particles of liquid when passing section m is 



LOSS OF ENEKGY. 



719 



equal to that when passing any other section, as n ; in steady 
flow without friction in rigid motionless vessels ; that is, the 
store of energy is constant. 

522. Bernoulli's Theorem with Friction, from the Standpoint of 
Energy. — Multiply each term in the equation of § 516 by Qy, 
as before, and denote a loss of head or height of resistance due 
to any cause by h r , and we have 

Qr^ + Qrf+Qy**. 

= Qr^+Qy p f+Qvz«-2 n Q r h r . . . (3) 



I v 2 
Each term Qyh r (e.g., Qy 4/* -y r— due to skin-friction in a 

d 2g 

long pipe, and Qy C E — - due to loss of head at the reservoir 
2g 

entrance of a pipe) represents a loss of energy, occurring between 
any position n and any other position m down-stream from n 9 
but is really still in existence in the form of heat generated by 
the friction of the liquid particles against each other or the 
sides of the pipes. 

As illustrative of several points in this connection, consider 
two short lengths of pipe in 
Fig. 575, A and B, one offering 
a gradual, the other a sudden, 
enlargement of section, but 
otherwise identical in dimen- 
sions. We suppose them to 
occupy places in separate lines 
of pipe in each of which a 
steady flow with full cross-sec- 
tions is proceeding, and so reg- 
ulated that the velocity and in- 
ternal pressure at n, in A, are 
equal respectively to those at n 




in B. Hence, if vacuum 



Fig. 575. 

piezometers be inserted at n, the 



720 MECHANICS OF ENGINEEEING. 

smaller section, the water columns maintained in them by the 
internal pressure will be of the same height, — , for both A 

r 

and B. Since at m, the larger section, the sectional area is the 
same for both A and B, and since F n in A = F n in B, so that 
Q A = Q By hence v m in A = v m in B and is less than v n . 

Now in B a loss of head occurs (and hence a loss of energy) 
between n and m, but none in A (except slight friction-head); 
hence in A we should find as much energy present at m as at 
n, only differently distributed among the three kinds, while at 
m in B the aggregate energy is less than that at n in B. 

As regards kinetic energy, there has been a loss between n 
and m in both A and B (and equal losses), for v m is less than 
v n . As to potential energy, there is no change between n and 
m either in A or B, since n and m are on a level. Hence if 
the loss of kinetic energy in B is not compensated for by an 
equal gain of pressure-energy (as it is in A), the pressure-head 

f^m) at m in B should be less than that (Sni\ at m in A. Ex- 
\y Ib \y I a 

periment shows this to be true, the loss of head being due to 
the internal friction in the eddy occasioned by the sudden en- 
largement ; the water column at m in B is found to be of a 
less height than that at m in A, whereas at n they are equal. 
(See p. 467 of article " Hydromechanics" in the Ency. Bri- 
tannica for Mr. Froude's experiments.) 

In brief, in A the loss of kinetic energy has been made up 
in pressure-energy, with no change of potential energy, but in 
B there is an actual absolute loss of energy = Qyh r , or 

v 2 
=53 QyZ -^-, suffered by the weight Qy of liquid. The value 
2g 

of C in this case and others will be considered in subsequent 

paragraphs. 

Similarly, losses of head, and therefore losses of energy, 

occur at elbows, sharp bends, and obstructions, causing eddies 

and internal friction, the amount of each loss for a given 

/y 2 'y 2 

weight, G, of water being = Gh r = GQ — - ; h r = £ — being 

«/ «/ 

the loss of head occasioned by the obstruction (§ 474). It is 



SUDDEN ENLAKGEMENTS IN PIPES. 721 

therefore very important in transmitting water through pipes 

for purposes of power to use all possible means of preventing 

disturbance and eddying among the liquid particles. E.g., 

sharp corners, turns, elbows, abrupt changes of section, should 

be avoided in the design of the supply-pipe. The amount of 

the losses of head, or heights of resistance, due to these various 

causes will now be considered (except skin -friction, already 

treated). Each such loss of head will be written in the form 

v 2 
C — , and we are principally concerned with the value of the 

abstract number C, or coefficient of resistance, in each case. 
The velocity v is the velocity, known or unknown, where the 
resistance occurs; or if the section of pipe changes at this 
place, then v = velocity on the down-stream section. The late 
Professor Weisbach, of the mining-school of Freiberg, Saxony, 
was one of the most noted experimenters in this respect, and 
will be frequently quoted. 

523. Loss of Head Due to Sudden (i.e., Square-edged) Enlarge- 
ment. Borda's Formula. — Fig. 576. An eddy is formed in the 

I _ -i angle with consequent loss of energy. Since 

j^JE ^M^JP* eacn particle of water of weight = G 1 , arriving 
#§p^=iyg^: with the velocity v 1 in the small pipe, may be 
* 1 1 U)T^^r t considered to have an impact against the base 

Fig. 576. of the infinitely great and more slowly moving 

column of water in the large pipe, and, after the impact, 
moves on with the same velocity, v 2 , as that column, just as 
occurs in inelastic direct central impact (§ 60), we may find 
the energy lost by this particle on account of the impact by 
eq. (1) of § 138, in which, putting M \ = 6r, -i- g, and M 2 = 
G 2 -±- g = mass of infinitely great body of water in the large 
pipe, so that M 2 = go , we have 

Energy lost by particle ■= G l i-i— — ^-, . . (1) 
and the corresponding 



Loss of head 



2<? 



722 



MECHANICS OF ENGINEERING. 



which, since F x v x = F 2 v a , may be written 

Loss of head in sudden enlargement = — i — 1 — l. . (2) 

That is, the coefficient C for a sudden enlargement is 

C = Sf - 1 )' " • • W 

F x and i^ 2 are the respective sectional areas of the pipes. Eq. 
(2) is Bordtfs Formula. 

[Note. — Practically, the now cannot always be maintained 
with full sections. In any case, if we assume the pipes to be 
running full (once started so), and on that assumption compute 
the internal pressure at F x , and find it to be zero or negative, 
the assumption is incorrect. That is, unless there is some 
pressure at F 1 the water will not swell out laterally to fill the 
large pipe. 

Example. — Fig. 577. In the short tube AB containing a 
sudden enlargement, we have given F 2 = F m = 6 sq. inches, 
F, = 4: sq. inches, and h = 9 feet. Re- 
quired the velocity of the jet at m (in 
the air, so that^? m -f- y = h = 34 ft.), if 
the only loss of head considered is that 
due to the sudden enlargement (skin- 
friction neglected, as the tube is short ; 
the reservoir entrance has rounded cor- 
ners). Applying Bernoulli's Theorem 
to m as down-stream section, and n in reservoir surface as up- 
stream position (datum level at m), we have 






r 



317i 



Fig. 577. 



^ +J + = + 5 + A-c|. . 



But, here, v 2 = v m ; 



From eq. (3) we have 



,(l + C)^ = A. 



. . (4) 

• • (5) 



z = a-iy = o.25, 



SUDDEN ENLAKGEMENT IN PIPE. 723 

and finally (ft., lb., sec.) 

v m = a/j^ V2 X 32.2 X 9 = 0.895 V2 X 32.2 X 9 
= 21.55 ft. per sec. 

(The factor 0.895 might be called a coefficient of velocity for 
this case.) Hence the volume of flow per second is 

Q = F m v m = ThX 21.55 = 0.898 cub. ft. per sec. 

We have so far assumed that the water fills both parts of the 
tube, i.e., that the pressure p 1 , at F l9 is greater than zero (see 
foregoing note). To verify this assumption, we compute p x 
by applying Bernoulli's Theorem to the centre of F x as down- 
stream position and datum plane, and n as up-stream position, 
with no loss of head between, and obtain 

^ + f + = + b + h-0 (6) 

But since F l v l = F,v, , we have 

v> = (f)V = (f)V 



2 

m 5 



and hence the pressure-head at F x (substituting from equations 
above) is 

^ + A_( f )^_ = 34+9-|. r ^ = 27feet, 

and .-. p x = H of 14.7 = 11.6 lbs. per sq. inch, which is 
greater than zero ; hence efflux with the tube full in both parts 
can be maintained under 9 ft. head. 

If, with F x and F 2 as before (and .*. £), we put p x = 0, and 
solve for A, we obtain h = 42.5 ft. as the maximum head 
under which efflux with the large portion full can be secured. 

524. Short Pipe, Square-edged Internally. — This case, already 



724 



MECHANICS OF ENGINEERING. 



a=90 



treated in §§ 507 and 515 (see Fig. 578 ; a repetition of 560), 
presents a loss of head due to the sudden enlargement from 
the contracted section at mf (whose sec- 
"l tional area may be put = OF, C being 
i an unknown coefficient, or ratio, of 
^ j contraction) to the full section F of 
lLlJT the pipe. From § 515 we know that 
^HJ the loss of head due to the short pipe 

is h r =C E p- (for a = 90°), in which 

Q E = 0.505 ; while from Borda's For- 
F 



^c^v 



iy. 



'*Z€.'r~& 



Fig. 578. 

mula, § 523, we have also Q E = 



-1 



Equating these, 



LCF 
we find the coefficient of internal contraction at mf to be 



C = 



1 + VZ E 1 + V.505 



= 0.584, 



or about 0.60 (compare with O — .64 for thin-plate contrac- 
tion, § 495). It is probably somewhat larger than this (.584), 
since a small part of the loss of head, A r , is due to friction at 
the corners and against the sides of the pipe. 

By a method similar to that pursued in the example of 
§ 523, we may show that unless h is less than 40 feet, about, 
the tube cannot be kept full, the discharge being as in Fig. 
551. If the efflux takes place into a " partial vacuum," this 
limiting value of h is still smaller. Weisbach's experiments 
confirm these statements (but those in the C. U. Hyd. Lab. 
seem to indicate that the limiting value for h in the first case 
is about 50 ft.). 






525. Diaphragm in a Cylindrical Pipe. — Fig. 579. The dia- 
phragm being of " thin plate," 
let the circular opening in it 
(concentric with the pipe) have 
an area = F, while the sectional ^_ 
area of pipe = F v . Beyond F, the fig. 579. 

stream contracts to a section of area = CF = F 1 , in enlarging 









SHOET PIPE. DIAPHEAGM IN PIPE. 



725 



from which to Jill the section F 2 , of pipe, a loss of head occurs 
which by Borda's Formula, § 523, is 



*•-&&-*) 



2<? 



where v 2 is the velocity in the pipe (supposed full). Of course 
F 1 (or OF) depends on F; but since experiments are necessary 
in any event, it is just as well to give the values of C itself, as 
determined by "Weisbach's experiments, viz. : 



F 
For^-= .10 


.20 


.30 


.40 


.50 


.60 


.70 


.80 


.90 


1.00 


C = 226. 


48. 


17.5 


7.8 


3.7 


1.8 


.8 


.3 


.06 


0.00 



By internal lateral filling, Fig. 580, the change of section 
may be made gradual and eddying 
prevented ; and then but little loss 
of head (and therefore little loss of 
energy) occurs, besides the slight 
amount due to skin-friction along 
this small surface. 




Fig. 580. 

On p. 467 of the article Hydromechanics 
in the Encyclopaedia Britannica may be found an account of 
experiments by Mr. Froude, illustrating this fact. 



526. "The Venturi Water-meter." — The invention bearing 
this name was made by Mr. Clemens Herschel (see Trans. Am- 
JSoc. Civ. Engineers^ for November 1887), and may be de- 
scribed as a portion of pipe in which a gradual narrowing of 
section is immediately succeeded by a more gradual enlarge- 
ment, as in Fig. 580 ; but the dimensions are more extreme. 
During the flow the piezometer-heights are observed at the 
three positions r, n_, and m (see below), and the rate of dis- 
charge may be computed as follows : Referring to Fig. 580, 
let us denote by r the (up-stream) position where the narrow- 
ing of the pipe begins, and by m that where the enlargement 
ends, while n refers to the narrowest section. F m = F r . 

Applying Bernoulli's Theorem to sections r and n, assuming 



726 MECHANICS OF ENGINEERING. 

no loss of head between, we have, as the principle of the ap- 
paratus, 

whence, since F r v r = F n v n , 



'•VrqfrV^-f^V^-# 



(2) 



in which <p represents the first radical factor. should differ 

F 
but little from unity with -£ small (and such was found to be 

the case by experiment). Its theoretical value is constant and 
greater than unity. In the actual use of the instrument the 

— and — are inferred from the observed piezometer- heights 

y r and y n (since — = y r -f- &, and « = y n -\- b, h being = 34 ft.), 

and then the quantity flowing per time-unit computed, from 
Q = F n v n , v n having been obtained from eq. (2). This pro- 
cess gives a value of Q about four per cent in excess of the 
truth, according to the second set of experiments mentioned 
below, if v n =35 ft. per sec. ; but only one per cent excess with 
v n = 5 or 6 ft. per sec. 

Experiments were made by Mr. Herschel on two meters of 
this kind, in each of which F n was only one ninth of F r , a 
ratio so extreme that the loss of head due to passage through 
the instrument is considerable. E.g., with the smaller appara- 
tus, in which the diameter at n was 4 in., the loss of head be- 
tween r and m was 10 or 11 ft., when the velocity through n 
was 50 ft. per sec, those at other velocities being roughly pro- 
portional to the square of the velocity. In the larger instru- 
ment d n was 3 ft., and the loss of head between r and m was 
much more nearly proportional to the square of the velocity 
than in the smaller. (E.g., with v n = 34.56 ft. per sec. the 
loss of head was 2.07 ft., while with v n = 16.96 ft. per sec. it 



SUDDEN DIMINUTION OF SECTION OF PIPE. 



727 



was 0.49 ft.) The angle of divergence was much smaller in 
these meters than that in Fig. 580. 



527. Sudden Diminution of Cross-section, Square Edges. — Fig. 
581. Here, again, the resistance is i 



~4^J^k J r^ 



due to the sudden enlargement from j^^^pYgub 

the contracted section to the full sec- -i_-J«$ §fcg 

tion F 2 of the small pipe, so that in r~~~-~^ y ^ \ 
the loss of head, by Borda's formula, fig. 581. 

v 2 VF 1 2 v 2 
h — r a — a 1 2 

the coefficient 



a) 



(2) 



depends on the coefficient of contraction (7; but this latter is 
influenced by the ratio of F 2 to F Q , the sectional area of the 
larger pipe, C being about .60 when F is very large (i.e., 
when the small pipe issues directly from a large reservoir so 
that F 2 : F practically = 0). For other values of this ratio 
"Weisbach gives the following table for O, from his own ex- 
periments : 



For i^ : F = .10 


.20 


.30 


.40 


.50 


.60 


.70 


.80 


.90 


1.00 


C= .624 


.632 


.643 


.659 


.681 


.712 


.755 


.813 


.892 


1.00 



O being found, we compute C from eq. (2) for use in 
eq.(l). 

528. Elbows. — The internal disturbance caused by an elbow, 
Fig. 582 (pipe full, both sides of elbow), occasions a loss of 
head 

a=c£,. •••(!) 

fig. 582. in which, according to Weisbach's experi- 

ments with tubes 3 centims., i.e. 1.2 in., in diameter, we may 
put 




728 



MECHANICS OF ENGINEEKING. 



For a = 20° 


40° 


60° 


80° 


90° 


100° 


110° 


120° 


130° 


140° 


C= .046 


.139 


.364 


.740 


.984 


1.26 


1.556 


1.86 


2.16 


2.43 



computed from the empirical formula ; 

C = .9457 sin 2 |a + 2.047 sin 4 \a ; 

v is the velocity in pipe ; a as in figure. For larger pipes C 
would probably be somewhat smaller ; and vice versa. 

If the elbow is immediately succeeded by another in the same 
plane and turning the same way, Fig. 583, the 
loss of head is not materially increased, since 
the eddying takes place chiefly 
in the further branch of the 
second elbow ; but if it turns 
in the reverse direction, Fig. 
584, but still in the same 
plane, the total loss of head is double that of 
one elbow ; while if the plane of the second is ~~\ 
to that of the first, the total loss of head is 1J 
times that of one alone. (Weisbach.) 



i 



^n&= 



ml 



Fig. 583. 




Fig. 584. 



529. Bends in Pipes of Circular Section. — Fig. 585. Weis- 
bach bases the following empirical 
formula for C, the coefficient of resist- 
ance of a quadrant bend in a pipe 
of circular section, on his own experi- 
ments and some of Dubuat's, viz. : 




for use in 



C = 0.131 + 1.847 (-)', . 



A r =C 



2P 



(1) 



where a = radius of pipe, r = radius of bend (to centre of 
pipe), and v — velocity in pipe ; h r = loss of head due to 
bend. 



ELBOWS AND BENDS IN PIPES. 



729 



It is understood that the portion BO of the pipe is kept full 
by the flow ; which, however, may not be practicable unless 
BO is more than three or four times as long as 
wide, and is full at the outset. A semicircular 
bend occasions about the same loss of head as a 
quadrant bend, but two quadrants forming a re- 
verse curve in the same plane, Fig. 586, occasion a 
double loss. By enlarging the pipe at the bend, 
or providing internal thin partitions parallel to the 
sides, the loss of head may be considerably dimin- 
ished. Weisbach gives the following table com- fig. 586. 
puted from eq. (1), but does not state the absolute size of the 
pipes. 




For -=.10 
r 


.20 .30 


.40 


.50 


.60 


.70 


.80 


.90 


1.0 


£=.131 


.138 


.158 


.206 


.294 


.440 


.661 


.977 


1.40 


1.98 



Accounts of many of Weisbach's hydraulic experiments are 
contained in the Cimlingenieur^ vols, ex, x, and xi. 

529a. Common Pipe-elbows. — Prof. L. F. Bellinger of Nor- 
wich University, Vermont, conducted a set of experiments in 
1887, when a student at Cornell, on the 
loss of head occasioned by a common el- 
bow (for wrought-iron pipe), whose longi- 
tudinal section is shown in Fig. 586a. 
The elbow served to connect at right 
angles two wrought-iron pipes having an 
internal diameter of 0.482 in. 
Fig. 586a. The internal diameter of the short bend 

or elbow was f in., and the radius of its curved circular axis 
(a quadrant) was £ in. Its internal surface was that of an 
ordinary rough casting. 

A straight pipe of the same character and size and 14 feet 
long was first used, and the loss of head due to skin-friction, 
(the only loss of head in that case) carefully determined for a 
range of velocities from 2 to 20 ft. per sec. 

Two lengths of similar pipe were then joined by the elbow 




730 



MECHANICS OF ENGINEERING. 



mentioned, forming a total length of 14 feet, and the total loss 
of head again determined through the same range of velocities. 
By subtraction, the loss of head due to the elbow was then 
easily found for each velocity, and assuming the form 



h=C 



~2g 



(1) 



for the loss of head, C was computed in each case. 

From Fig. 586a it is seen that the stream meets with a sud- 
den enlargement and a sudden diminution, of section, as well 
as with the short bend ; so that the disturbance is of a rather 
complex nature. 

The principal results of Prof. Bellinger's experiments, after 
the adjustment of the observed quantities by " least squares,'' 
were found capable of being represented fairly well by the 
formula 

C = 0.621 + [2 W - 1] X 0.0376, . . . (2) 

where n = [veloc. in pipe in ft. per sec] -=- 5. The following 
table was computed from eq. (2) (where v is in ft. per second) : 



1) = 


2 


4 


6 


8 


10 


12 


14 


16 


18 


20 


c= 


.633 


.649 


.670 


.697 


.734 


.782 


.845 


.929 


1.039 


1.185 



530. Valve-gates and Throttle- valves in Cylindrical Pipes. — 

Adopting, as usual, the form 






(i) 



for the loss of head due to a valve-gate, Fig. 587, or for a 

"T , —J 




Fig. 587. Fig. 588. 

throttle-valve. Fig. 588, each in a definite position, Weisbach's 



VALYE-GATE. THROTTLE-YALVE. 



731 



experiments furnish us with a range of values of C in the case 
of these obstacles in a cylindrical pipe 1.6 inches in diameter, 
as follows (for meaning of s, d, and a, see figures, v is 
the velocity in the full section of pipe, running full on both 
sides.) 



Valve-gate. 


Throttle-valve. 


S 
d 


c 


a 








5° 


.24 


1.0 


.00 


10° 


.52 


I 


.07 


15° 

20° 


.90 
1.54 


f 


.26 


25° 


2.51 




30° 


3.91 


f 


.81 


35° 


6.22 






40° 


10.8 


4 
5 


2.06 


45° 


18.7 


1 


5.52 


50° 
55° 


32.6 

58.8 


2 


17 00 


60° 


118.0 


¥ 




65° 


256.0 


i 


97.8 


70° 


751. 











531. Examples involving Divers Losses of Head. — "We here 
suppose, as before, that the pipes are full during the flow. 
Practically, provision must be made for the escape of the aiz 
which collects at the high points. If this air is at a tension 
greater than one atmosphere, automatic air-valves will serve to 
provide for its escape ; if less than one atmosphere, an air- 
pump can be used, as in the case of a siphon used at the 
Kansas City "Water Works. (See p. 346 of the Engineering 
News for November 1887.) 



9C D 



X^rE 






—Tc 



J 2 =20 



B— — Z*=-. 



■%~ 



m 



Fig. 589. 



Example 1. — Fig. 589. "What head, = A, will be required 
to deliver -J U. S. gallon (i.e. 231 cubic inches) per second 



732 MECHANICS OF ENGINEERING. 

through the continuous line of pipe in the figure, containing two 
sizes of cylindrical pipe (d = 3 in., and d 2 = 1 in.), and two 
90° elbows in the larger. The flow is into the air at m, the 
jet being 1 in. in diameter, like the pipe. At E, a = 90°, and 
the corners are not rounded ; at K, also, corners not rounded. 
Use the ft.-lb.-sec system of units in which g = 32.2. 

Since Q = -J- gal. = \ • y 2 ^ = .0668 cub. ft. per sec, and 
therefore the velocity of the jet 

v m = i> 2 = Q-± \n{-^f = 12.25 ft. per sec; 

hence the velocity in the large pipe is to be v = (-J) 8 ^ = 1.36 
ft. per sec. From Bernoulli's Theorem, we have, with m as 
datum plane, 

v 2 v 2 I v 2 

involving six separate losses of head, for each of which there 
is no difficulty in finding the proper £ or/*, since the velocities 
and dimensions are all known, by consulting preceding para- 
graphs. (Clean iron pipe.) 

From § 515, table, for a = 90° we have . . . C* = 0.505 
" § 517, for d = 3in.,and0 o =1.36ft.persec.,/ o = .00725 
« « "4 = 1 in., and^ ^12.25 " " /, = .00613 

« § 528 (elbows), for a = 90° .... C el _ =0.984 
" § 527, for sudden diminution at K we have 
[since F, -r- F Q = V -f- 3 a = 0.111, .-. (7=0.625] 

Q K = [— - lV = 0.360. 

* K V.625 I 

Solving the above equation for h, then, and substituting 
above numerical values (in ft.-lb.-scc.-system), we have (noting 
that v m = 4 , and v Q = ^ a ) 

h = (1^ 5 2 fl + (tf (.505 + 4 X •° ^ 5 X 5 ° + 2 X .984) 
64.4 L ^ I- ' 

12 -* 



EXAMPLES : WITH LOSSES OF HEAD. 



783 



i.e., 

h = Q^p- [l +(.00623 +.07160 + .0243)+(.360+5.8848~l; 
64.4 L 

.-. h = 2.323 X 7.3469 = 17.09 ft— An*. 

It is here noticeable how small are the losses of head in the 
large pipe, the principal reason of this being that the velocity 
in it is so small (v = only 1.36 ft. per sec), and that in gen- 
eral losses of head depend on the square of the velocity 
(nearly). 

In other words, the large pipe approximates to being a reser- 
voir in itself. 

With no resistances & head h = -y TO 2 -r- 2g = 2.32 ft. would be 
sufficient. 

Example 2. — Fig. 590. "With the valve-gate I 7 ' half raised 
(i.e., s = id in Fig. 587), required the volume delivered per 
second through the clean pipe here shown. The jet issues 



<£%•>.- .-, 




Fig. 590. 

from a short straight pipe, or nozzle (of diameter d^ = 1J in.) 
inserted in the end of the larger pipe, with the inner corners 
not rounded. Dimensions as in figure. Radius of each bend 
= r = 2 in. The velocity v m of the jet in the air = velocity 
v 2 in the small pipe ; hence the loss of head at iTis 






r -. 



Now v m is unknown, as yet ; but v , the velocity in the large 
pipe, is —v m pp ; i.e., v = -^v m . From Bernoulli's The- 



784 MECHANICS OF ENGINEEKING. 

orem (pi as datum level) we obtain, after transposition, 

h ^ +z 4 +z 4 + " ZB % +if 'i v i + ^4- (1) 

Of the coefficients concerned, f alone depends on the un- 
known velocity v . For the present [first approximation], 

pnt ■..:•/.= .006 

From § 515, with a = 90°, Q E = .505 

From § 517, valve-gate with s = %d, C F = 2.06 

From § 529, with a : r = 0.5, Q B = 0.294 

While at IT, from § 527, having 

(F 9 :F ) = (iy:V = ft= 0.562; 

we find from table, C = 0.700 

and .-. C*=(^-l) 2 =(0.428) 2 . . . . i.e., C^= 0.183 
Substituting in eq. (1) above, with v* = (-j^) 2 v m 2 , we have 
/ I , 

in which the first radical, an abstract number, might be called 
a coefficient of velocity, 0, for the whole delivery pipe ; and 
also, since in this case Q, = F m v m = F^v 2 , may be written 
Q = jxF 2 V%gh y it may be named a coefficient of effiux, fx. 
Hence 



Vm = A / Q , r~ 4 x 006 x 80~l V 2 x 32.2 x 25; 

A / l-h^j-r .505 + 2.06 r 2 x .294 + 4x • U0bx8U +.183 

.-. v m = 0.421 V2gh = 0.421 4/2x32.2x25 = 16.89 ft. per see 

(The .421 might be called a coefficient of velocity.) The 
volume delivered per second is 

Q = \nd* v m = iTti&y 16.89 = .207 cub. ft. per sec. 

(As the section of the jet F m = F 2 , that of the short pipe or 
nozzle, we might also say th at .42 1 = jjl = coefficient of efflux % 
for we may write Q == p.F 2 V2gh, whence jj. = .421.) 



FLOW THEOUGH SIPHONS. 735 

532. Siphons.— In Fig. 532, § 490, the portion HN 2 is 
above the level, BC, of the surface of the water in the head 
reservoir BL, and yet under proper conditions a steady flow- 
can be maintained with all parts of the pipe full of water, in- 
cluding HJV 2 0, If the atmosphere exerted no pressure, this 
would be impossible ; but its average tension of 14.7 lbs. per 
sq. inch is equivalent to an additional depth of nearly 34 feet 
of water placed upon BC. With no flow, or a very small 
velocity, the pipe may be kept full if JV 2 is not more than 
33 or 34 feet above BO; but the greater v 2 , the velocity of 
flow at iV^, and the greater and more numerous the losses 
of head between L and N % , the less must be the height of N^ 
above BO for a steady flow. 

The analytical criterion as to whether a flow can be main- 
tained or not, supposing the pipe completely filled at the out- 
set, is that the internal pressure must be > at all parts of 
the pipe. If on the supposition of a flow through a pipe of 
given design the pressure^? is found < 0, i.e. negative, at any 
point [iV^ being the important section for test] the supposition 
is inadmissible, and the design must be altered. 

For example, Fig. 532, suppose LNJSf^ to be a long pipe of 
uniform section (diameter = d, and length = Z), and that under 
the assumption of filled sections we have computed v At the 
velocity of the jet at J¥ A ; i.e., 



V 



J^gh (1) 



To test the supposition, apply Bernoulli's Theorem to the 
surface BC and the point iV 2 where the pressure is^> 2 , velocity 
v£= v A , since we have supposed a uniform section for whole 
pipe), and height above BC=h 2 . Also, let length of pipe 



LN X HN % = Z a . Whence we have 



[BC being datum plane.] 



736 MECHANICS OF ENGINEERING. 

Solving for ^— , we have 
Y 

*=,«*£■*- [a, + |.(i + C £+ 4/|)]. . . (3) 

We note, then, that for^> 2 to be > 0, 

A a ^^<[34/^-|-C^-4/^].. (4> 

In the practical working of a siphon it is found that atmos- 
pheric air, previously dissolved in the water, gradually collects 
at JV a , the highest point, during the flow and finally, if not re- 
moved, causes the latter to cease. See reference below. 

One device for removing the air consists in first allowing it 
to collect in a chamber in communication with the pipe be- 
neath. This communication is closed by a stop-cock after the 
water in it has been completely displaced by air. Another 
stop-cock, above, being now opened, water is poured in to re- 
place the air, which now escapes. Then the upper stop-cock is 
shut and the lower one opened. The same operation is again 
necessary, after some hours. 

On p. 346 of the Engineering News of November 1887 may 
be found an account of a siphon which has been employed since 
1875 in connection with the water-works at Kansas City. It 
is 1350 ft. long, and transmits water from the river to the 
artificial " well " from which the pumping engines draw their 
supply. At the highest point, which is 16 ft. above low- water 
level of the river, is placed a " vacuum chamber " in which the 
air collects under a low tension corresponding to the height, 
and a pump is kept constantly at work to remove the air and 
prevent the " breaking" of the (partial) vacuum. The diam- 
eter of the pipe is 24 in., and the extremity in the " well " dips 
5 ft. below the level of low water. See Trautwine's Pocket- 
book, for an account of Maj. Crozet's Siphon. 

532a. Branching Pipes. — If the flow of water in a pipe is 
caused to divide and pass into two others having a common 



BRANCHING PIPES. 73? 

junction with the first, or vice versa ; or if lateral pipes lead 
out of a main pipe, the problem presented may be very com- 
plicated. As a comparatively simple instance, let us suppose 
that a pipe of diameter d and length I leads out of a reservoir, 
and at its extremity is joined to two others of diameters d t and 
d 2 and lengths l x and Z 2 respectively, and that the further extrem- 
ities of the latter discharge into the air without nozzles under 
heads h 1 and A 2 below the reservoir surface. Call these two 
pipes Nos. 1 and 2. That is, the system forms a Y in plan. 

Assuming that all entrances and junctions are smoothly 
rounded, so that all loss of head is due to skin-friction, it is re- 
quired to find the three velocities of flow, v, v l9 and v 3 , in the 
respective pipes. First applying Bernoulli's Theorem to a 
stream-line from the reservoir surface through the main pipe 
to the jet at the discharging end of pipe No. 1, we have 

2g l J d 2g J d, 2g ' w 

and similarly, dealing with a stream-line through the main 
pipe and No. 2, 

2g 2 J d 2g ^ d % ' 2g ' * * ' (2) 

while the equation of continuity for this case is 

ixcTv = i7id 1 \ + i7rd a \ (3) 

From these three equations, assuming/* the same in all pipes 
as a first approximation, we can find the three velocities (best 
by numerical trial, perhaps) ; and then the volume of discharge 
of the system per unit of time 

Q = i*tfv (4) 

533. Time of Emptying Vertical Prismatic Vessels (or Inclined 
Prisms if Bottom is Horizontal) under Variable Head. 

Case I. Through an orifice or short pipe in the bottom and 
opening into the air. — Fig. 591. As the upper free surface, 



738 MECHANICS OF ENGINEERING. 

of area = F' , sinks, F' remains constant. Let z == head of 
water at any stage of the emptying ; it = z at the outset, and 
= when the vessel is empty. At any 
instant, Q, the rate of discharge (= vol- 
ume per time-unit) depends on z and is 



— J__| -=: 

—s r — 



f- Q = M FV2gz, . . . (1) 



\( where /x = coefficient of efflux = <pC = 

FlG - 591 - coefficient of velocity X coefficient of con- 

traction [see § 495, eq (3)]. We here suppose F' so large 
compared with F, the area of the orifice, that the free surface 
of the water in the vessel does not acquire any notable velocity 
at any stage, and that hence the rate of efflux is the same at 
any instant, as for a steady flow with head — z and a zero 
velocity in the free surface, jjl is considered constant. From 
(1) we have 

dV = (vol. discharged in time dt) = Qdt = jxFVtyz dt, . (2 

But this is also equal to the volume of the horizontal lamina, 

F'dz, through which the free surface has sunk in the same 

time dt 

— —F' 

... _ F'dz = jiFVtysMt; .\ dt = ,— z~*dz. . (3) 

We have written minus F'dz because, dt being an increment, 
dz is a decrement. To reduce the depth from z (at the start, 
time = t = zero) to z n , demands a time 

H= 1 ~ 1 = P n z-*dz= Jt^—lzi-ztf; (4) 

whence, by putting z n = 0, we have the time necessary to 
empty the whole prism 



c 



2F'z* _ %F'z 2 X volume of vessel . „. 

" pFVty ~~ fxFV2gz, ~ mifcial rate of discharge ' 



TIME OF EMPTYING VESSELS. 



739 



that is, to empty the vessel requires double the time of dis- 
charging the same amount of water if the vessel had been kept 
full (at constant head = z = altitude of prism). 

To fill the same vessel through an orifice in the bottom, the 
flow through which is supplied from a 
body of water of infinite extent hori- 
zontally, as with the (single) canal lock 
of Fig. 592, will obviously require the 
same time as given in eq. (5) above, 
since the effective head z diminishes 
from z to 0, according to the same law. 

Example. — What time will be needed 
to empty a parallelopipedical tank (Fig. 591) 4 ft. by 5 ft. in 
horizontal section and 6 ft. deep, through a stop-cock in the 
bottom whose coefficient of efflux when fully open is known 
to be /a. = 0.640, and whose section of discharge is a circle of 
diameter — \ in. ? From given dimensions F' = 4 X 5 = 20 
sq. ft., while z = 6 ft. Hence from eq. (5) (ft.-lb.-sec.) 




Fig. 592. 



2 X 20 X V6 



time of \ 

emptying \ 0.64 X \n(jkif V2 X 32.2 



13980 seconds 

— q hours g g min. A sec. 



Case II. Two communicating prismatic vessels. Required 
the time for the water to come to a common level OJV, Fig. 
593, efflux taking place through a small 
orifice, of area == F, under water. At 
any instant the rate of discharge is 

Q = jxFVfyz~, 

as before, z = difference of level. Now 
if F' and F" are the horizontal sectional 
fig. 593. areas of the two prismatic vessels (axes 

vertical) we have F'x = F"y, and hence 2, which = x -{-y y 

= x + (F' + F")x; 



"7" 


- 7 7-r 

y-j-j-- 


|AJ. 


IpfEr 



X = 






and dx — 



dz 



1 + 



El 

JT" 



740 MECHANICS OF ENGINEEBING. 

As before, we have 

F'F" z~Wz 
- F'dx = piFV2g zUt, or dt=- • , , - _ ,— . 

Hence, integrating, the time for the difference of level to. 
change from z to z n 



ZF'F' 



*<? — Zy? 



F f +F ,r }xFV¥g' 
and by making z n — in (6), we have the 

time of coming to a common 



(6) 



Uwl =TTF*\]nr\/tg (7 > 



Algebraic Example. — In the double lock in Fig. 594, let 
U be full, while in L" the water stands at a level MN the 

same as that of the tail- 



— L- 



A- S\ 






water. F' and F" are the 
horizontal sectional areas of 
the prismatic locks. Let 
the orifice, 0, between 
them, be at a depth = h t 
below the initial level KE 
of Z', and a height = A 2 
above that, MN, of Z". 
The orifice at 0, area = F, being opened, efflux from L' be- 
gins into the air, and the level of L" is gradually raised from 
MN to OD, while that of L' sinks from KE to J.i? a distance 
= a, computed from the relation vol. F'a — vol. F"\, and 
the time occupied is [eq. (4)] 

2F' 



Fig. 594. 



& = 



MEVty 



==[ VA 2 — VA, — a]. 



(8) 



As soon as is submerged, efflux takes place under water, and 
we have an instance of Case II. Hence the time of reaching 
a common level (after submersion of 0) (see eq. 7) is 



2F'F" 

2_ jjlF{F'-\-F") 



V 



and the total time is = t l -f- 1 2 , with a 



A, — a 
F"h, 



(»> 



J?". 



TIME OF EMPTYING VESSELS. 



741 



Case III. Emptying a vertical prismatic 
rectangular "notch" in the side, or over- 
fall. — Fig. 595. As before, let even the 
initial area (= zfi) of the notch be small 
compared with the horizontal area F' of 
tank. Let z = depth of lower sill of notch 
below level of tank surface at any instant, 
and b — width of notch. Then, at any in- 
stant (see eq. 10, § 504), 



vessel through a 




Fig. 595. 



Bate of disch. (vol.) = Q = %j* bz V2gz = %jrf> V2gzK 

.\ vol. of disch. in dt = \pb V2g z%dt, 

and putting this = — F r dz = vol. of water lost by the tank 
in time dt, we have 



dt= — -— — -s~t^; 



whence 



2/iJ V2g 



U 2/xb V2gjz 



3 F 



* r-n 



-* 



i.e., 



n^JlLrJ^-Js],.. . . (10) 

Lo MbV2gLVz n Vz J 

as the time in which the tank surface sinks from a height z 
above sill to a height z n above sill. If we inquire the time t' 
for the water to sink to the level of the sill of the notch we 
put z n = zero, whence t' = infinity. As explanatory of this 
result, note that as z diminishes not only does the velocity of 
flow diminish, but the available area of efflux (= zb) also grows 
less, whereas in Cases I and II the orifice of efflux remained 
of constant area = F. 

Eq. (10) is applicable to the waste-weir of a large reservoir 
or pond. 

534. Time of Emptying Vessels of Variable Horizontal Sec- 
tions. — Considering regular geometrical forms first, let us take 



742 



MECHANICS OF ENGINEERING. 



Case L 



Wedge-shaped vessel, 
neath, 



edge horizontal and under- 
m orifice F in the edge, so that 
s, the variable head, is always the 
altitude of a triangle similar to the 
section ABC of the body of water 
when efflux begins. At any instant 
during the efflux the area, S, of 
the free surface, variable here, 
takes the place of F' in eq. (3) of 
§ 533, whence we have, 

—Sz-*dz 
for any case of variable free surface, dt = . . (11) 

/xFV2g 

In the present case S = ul, and from similar triangles 

u : z : : b : z Q ; 
whence 

and 




Fig. 596. 



S=blz 



dt = 



— blzMz 



t = -== / z?dz = 

U M Fz V2gJ z n 



\bl 



2„* 



■/], ■ (12) 



*o = 



(13) 



M Fz V2gJ z n }*Fz V2g 

and hence the time of emptying the whole wedge, putting 
z n = 0, is 

4 iblz _ 4 Vol. of wedge 

3 uF V%qz 3 initial rate of discharge 

i.e., |- as long as to discharge the same volume of water under 
a constant head = z . This is equally true if the ends of the 
wedge are oblique, so long as they are parallel. 
Case II. Right segment of paraboloid of 
revolution. — Fig. 597. Axis vertical. Ori- 
fice at vertex. Here the variable free surface 
has at any instant an area, = S, = nu*, u be- 
ing the radius of the circle and variable. 
From a property of the parabola 



w 



S s± n b*z -J- s , 




Fig. 537. 



TIME OF EMPTYING VESSELS. 743 

and hence, from eq. (11), 

dt=—— — — -; 

Lo pFe>V5gJ*n ^ M Fz V2g "" 

whence, putting z n = 0, we have the time of emptying the 
whole vessel 

_ 4 7tb' i ^z Q _ 4 total vol. ( . 

~~ 3 j*FV2gz 3 initial rate of disch? 

same result as for the wedge, in Case I ; in fact, it applies to 
any vessel in which the areas of horizontal sections vary 
directly toith their heights above the orifice. 

Case III. Any pyramid or cone ; vertex down y small ori- 
fice in vertex. — Fig. 598. Let area of the s >r—~~^^ 
base = S , at upper edge of vessel. At f^ZJ ^^ 
any stage of the flow 8 = area of base of , / ,IZY^/ 
pyramid of water. From similar pyra- site4^#^ 
mids W&7 ? 

S - o S::z 9 *:s 9 ; .-. S = ^z% $*' JL ~ 

3 Fig. 598. 

and [eq. (11)] 

dt=-^\ — X -^ =z %dz, 

z* txFVZg 

whence (z n = 0) the time of emptying the whole vessel is 

-I s 



i= * _. ft* 2 ^ 



/xFz* V2g Jo 5 M^z; \/2g 

__ 6 Total volume /1 ^ 

' ° 5 initial rate of disch.' ' * 

Case IV. Sphere. — Similarly, we may show that to empty 



744 



MECHANICS OF ENGINEEKING. 



a sphere, of radius = r, through a small orifice, of area = F, 
in lowest part, the necessary time is 



. _ 16 71T % 



8 



Vol. 



15 ^F Vgr & init. rate of disch. 



535. Time of Emptying an Obelisk-shaped Vessel. — (An obe- 
lisk may be defined as a solid of six plane faces, two of which 
are rectangles in parallel planes and with sides respectively 
parallel, the others trapezoids; a frustum of a pyramid is a 
particular case.) 

A volume of this shape is of common occurrence ; see Fig. 
599. Let the altitude = A, the two rectangular faces being 
horizontal, with dimensions as in figure. By drawing through 

F, G, and H right lines par- 
allel to EC, to cut the upper 
base, we form a rectangle 
KLMC equal to the lower 
base. Produce ML to P and 
KL to N, and join FG and 
NG. We have now sub- 
divided the solid into a paral- 
lelopiped KLMC- EHGF, 
a pyramid PBNL - G, and 
two wedges, viz. APLK-HG and LNDM-FG, with 
their edges horizontal ; and may obtain the time necessary to 
empty the whole obelisk-volume by adding the times which 
would be necessary to empty the individual component vol- 
umes, separately, through the same orifice or pipe in the bot- 
tom plane EG. These have been already determined in the 
preceding paragraphs. The dimensions of each component 
volume may be expressed in terms of those of the obelisk, and 
all have a common altitude = h. 

Assuming the orifice to be in the bottom, or else that the 
discharging end of the pipe, if such is used, is in the plane of 
the bottom FG, we have as follows, F being the area of dis- 
charge : 




Fig. 599. 



TIME OF EMPTYING KESERVOIRS. 745 

Time to empty the parallelopiped) y — __^A_ j-r /-,\ 
separately would he {Case /, § 533) j * ' uF V%q 

Time to empty the two 1 g j „ _ () + ^ _ . ) 

For the pyramid ) f _2 (£ — ?J (ft — £,) /? - . 

(6bw 2Z7, § 584) } ' • - Cs -5 - jjpyfy VA< • * W 

Hence to empty the whole reservoir we have 

i.e., 

< = [8M + &J, + 2M, + »,q ^ 2 ^^_, . . . (4) 

Example. — Let a reservoir of above form, and with o = 50 ft., 
2 = 60 ft., 5, = 10 ft., l x = 20 ft., and depth of water h = 16 
ft., be emptied through a straight iron pipe, horizontal, and 
leaving the side of the reservoir close to the bottom, at an angle 
a — 36° with the inner plane of side. The pipe is 80 ft. long 
and 4 inches in internal diameter ; and of clean surface. The 
jet issues directly from this pipe into the air, and hence 
F=l7t(^f sq. feet. To find /*, the "coefficient of efflux" 
(= 0, the coefficient of velocity in this case, since there is no 
contraction at discharge orifice), we use eq. (4) (the first radical) 
of § 518, with/ approx. = .006, and obtain 



/ i r 






V 1+W-Vi \/ l+.896+i><^^ : 



(~N\B. Since the velocity in the pipe diminishes from a 
value 

v = .361 V2g X 16 = 11.6 ft. per sec. 

at the beginning of the flow to v = zero at the close,/" = .006 
is a reasonably approximate average with which to compute 
the average above ; see § 517. 



746 



MECHANICS OF ENGINEEKINGr. 



Hence from eq. (4) of this paragraph (ft.-lb.-sec. system) 

_ [3 X 50 X 60 + 8 X 10 X 20 + 2(50 X 20+10 X 60)]2 VU 



15 X 0.361 X 



3©"* 



X32.2 



o Q11 a „ ftrt q i k • j Probably within 2 or 3% of 
- 29110 sec. = 8 hrs. 5 mm. j J ^ truth< 

536. Time of Emptying Reservoirs of Irregular Shape. Simp- 
son's Rule. — From eq. (11), § 534, we have, for the time in 
which the free surface of water in a vessel of any shape what- 
ever sinks through a vertical distance =dz, 



— £z~Hz . 
at = - ■— - — , whence 



pFVty 



"* = *« I 

time = =^ 

■*=z MFV2g 



£z~Uz, . . (1) 



where £ is the variable area of the free surface at any in- 
stant, and z the head of water at the same instant, efflux 
proceeding through a small orifice (or extremity of pipe) of 
area — F. If £ can be expressed in terms of s, we can in- 
tegrate eq. (1) (i.e., provided thai £z~t has a known anti- 
derivative) ; but if not, the vessel or reservoir being irregular 
in form, as in Fig. 600 (which shows a pond whose bottom 
has been accurately surveyed, so that we know the value of £ 
for any stage of the emptying), we can still get an approximate 

solution by using Simpson's 
Rule for approximate inte- 
gration. Accordingly, if we 
inquire the time in which 
the surface will sink from 
to the entrance i^of the pipe 
in Fig. 600 (any point n ; at 
E. or short of that), we 
divide the vertical distance 
from to n (4 in this figure) into an even number of equal 
parts, and from the known form of the pond compute the area 
£ corresponding to each point of division, calling them £ Q , £ l9 
etc. Then the required time is approximately 




Fig. 600. 



TIME OF EMPTYING POND. 747 

^2 ~4 *?|,_2 "'M. I 

Example. — Fig. 600. Suppose we have a pipe Em of the 
same design as in the example of § 535, and an initial head of 
z Q = 16 ft., so that the same value of ju, = .361, may be used. 
Let z n — z Q = 8 feet, and divide this interval (of 8 ft.) into 
four equal vertical spaces of 2 ft. each. If at the respective 
points of division we find from a previous survey that S = 
400000 sq. ft., S, = 320000 sq. ft., S, = 270000 sq. ft., £ 3 = 
210000 sq. ft., and S, = 180000 sq. ft. ; while n = 4, pi = .361, 
and the area F — J7r(£) a = .0873 sq. ft., we obtain (ft., lb., sec.) 



G 



t = fe 



o 0.361 X .0873 V 2 X 32.2 X 3 X 4 



" 400000 4 X 320000 
- 1/16 Vl? 



2 X 270000 , 4 X 210000 , 180000" 



i/12 1/10 VS 



= 2444000 sec. 
= 28 d - 6 h - 53 m - 20 s - 



The volume discharged, V, may also be found by Simpson's 
Rule, thus: Since each infinitely small horizontal lamina has 
a volume 

dV=-Sde, /. rV= t £sdz, 
-o 

or, approximately, 

Hence with n = 4 we have (ft., lb., sec.) 

C F - fe 8 [ 40000 ° + M +SS } +^ 270000 

+ 180000~| = 2,160,000 cub. ft. 



748 



MECHANICS OF ENGINEEKING. 



537. Volume of Irregular Reservoir Determined by Observing 
Progress of Emptying. — Transforming eq. (11), § 534, we have 

Sdz = - }xF Vty z*dt. 

But Sdz is the infinitely small volume d V oi water lost by 
the reservoir in the time dt, so that the volume of the reser- 
voir between the initial and final (0 and n) positions of the 
horizontal free surface (at beginning and end of the time t n ) 
may be written 



V 



= l*FV¥g I s*dt. 



(1) 



This can be integrated approximately by Simpson's Kule, if 
the whole time of emptying, = t n , be divided into an even 

......;:-..-..:„ ._„*,-- .,--.,,.... . number of equal 

parts, and the values 

z o ? z x i z i ? e ^ c -j °f the 
head of water noted 
at these equal inter- 
vals of time (not of 
vertical height). The 
corresponding sur- 
face planes will not 
Whence for the particular case 




Fig. 601. 



be equidistant, in general, 
when n = 4 (see Fig. 601) 

ry = ^vm -o) w 4 2 ^ aJ , . _ _ (2) 

L o o X 4 



CHAPTEE VII. 




HYDRODYNAMICS (Continued)— STEADY FLOW OF WATER IN 
OPEN CHANNELS. 

538. Nomenclature. — Fig. 602. When water flows in an 
open channel, as in rivers, canals, mill-races, water-courses, 
ditches, etc., the bed 
and banks being rigid, 
the upper surface is 
free to conform in 
shape to the dynamic 
conditions of each case, 
which therefore regu- 
late to that extent the 
shape of the cross-sec- 
tion. 

In the vertical trans- FlG - 602 - 

verse section A C in figure, the line A C is called the air-profile 
(usually to be considered horizontal and straight), while the 
line ABC, or profile of the bed and banks, is called the wetted 
perimeter. It is evident that the ratio of the wetted perimeter 
to the whole perimeter, though never < -J, varies with the 
form of the transverse section. 

In a longitudinal section of the stream, EFGH, the angle 
made by a surface filament .ET^with the horizontal is called 
the slope, and is measured by the ratio s = h : I, where I is the 
length of a portion of the filament and h = the/all, or vertical 
distance between the two ends of that length. The angle be- 
tween the horizontal and the line HG along the bottom is not 
necessarily equal to that of the surface, unless the portion of 
the stream forms a prism ; i.e., the slope of the bed does not 
necessarily — s — that of surface. 

Examples. — The old Croton Aqueduct has a slope of 1.10 
ft. per mile ; i.e., s == .000208. The new aqueduct (for New 

749 



750 MECHANICS OF ENGINEEEING. 

York) has a slope s = .000132, with a larger transverse section. 
For large sluggish rivers s is much smaller. 

539. Velocity Measurements. — Various instruments and 
methods may be employed for this object, some of which are 
the following : 

Surface-floats are small balls, or pieces of wood, etc., so 
colored and weighted as to be readily seen, and still but little 
affected by the wind. These are allowed to float with the cur- 
rent in different parts of the width of the stream, and the sur- 
face velocity c in each experiment computed from c = Z -=- t, 
where I is the distance described between parallel transverse 
alignments (or actual ropes where possible), whose distance 
apart is measured on the bank, and t — the time occupied. 

Double-floats. Two balls (or small kegs) of same bulk and 
condition of surface, one lighter, the other heavier than water, 
^•^i,v9%:^/v.;^;A.,:;vv?... are united by a slender chain, their 
weights being so adjusted that the 
light ball, without projecting notably 
above the surface, buoys the other 
ball at any assigned depth. Fig. 603. 
It is assumed that the combination 
moves with a velocity c', equal to the 
Fig. 603. arithmetic mean of the surface veloc- 

ity c of the stream and that, c, of the water filaments at the 
depth of the lower ball, which latter, c, is generally less than 
c . That is, we have 

c' = i(e + o) and .-. c = 2c r - c . . . . (1) 

Hence, c Q having been previously obtained, eq. (1) gives the 
velocity c at any depth of the lower ball, c' being observed. 

The floating staff is a hollow cylindrical rod, of adjustable 
length, weighted to float upright with the top just visible. Its 
observed velocity is assumed to be an average of the velocities 
of all the filaments lying between the ends of the rod. 

Woltmanrts Mill ; or Tachometer ; or Current-meter, Fig. 
604, consists of a small wheel with inclined floats (or of a small 




CURRENT-METERS, ETC. 



751 




Fig. 604. 



" screw-propeller" wheel S) held with its plane 1 to the cur- 
rent, which causes it to re- 
volve at a speed nearly pro- 
portional to the velocity, c, 
of the water passing it. 
By a screw-gearing W on 
the shaft, connection is 
made with a counting ap- 
paratus to record the num- 
ber of revolutions. Some- 
times a vane B is attached, 
to compel the wheel to face 
the current. It is either 
held at the extremity of a pole or, by being adjustable along 
a vertical staff fixed in the bed, may be set .at any desired depth 
below the surface. It is usually so designed as to be thrown 
in and out of gear by a cord and spring, that the time of mak- 
ing the indicated number of revolutions may be exactly noted. 
By experiments in currents of known velocities a table or 
formula can be constructed by which to interpret the indica- 
tions of any one instrument ; i.e., to find the velocity c of the 
current corresponding to an observed number of revolutions 
per minute. 

A peculiar form of this instrument has been recently in- 
vented, called the Ritchie-Haskell Direction-current Meter, 
for which the following is claimed : " This meter registers 
electrically on dials in boat ^a^^m 

from which used, the direction 
and velocity, simultaneously, 
of any current. Can be used 
in river, harbor, or ocean cur- 
rents/' 

Pitotfs Tube consists in prin- 
ciple of a vertical tube open 
above, while its lower end, also 
open, is bent horizontally up- 
stream; see A in figure. After FlG . 6 05. 
the oscillations have ceased, the water in the tube remains 




752 MECHANICS OF ENGINEEKING. 

stationary with its free surface a height, h\ above that of the 
stream, on account of the continuous impact of the current 
against the lower end of the column. By the addition of 
another vertical tube (see B in figure) with the face of its 
lower (open) end parallel to the current (so that the water- 
level in it is the same as that of the current), both tubes being 
provided with stop-cocks, we may, after closing the stop- 
cocks, lift the apparatus into a boat and read off the height hi 
at leisure. We may also cause both columns of water to 
mount, through flexible tubes, into convenient tubes in the 
boat by putting the upper ends of both tubes in communica- 
tion with a receiver of rarefied air, and thus watch the oscilla- 
tions and obtain a more accurate value of h\ [See Yan Nos- 
trand's Mag. for Mar. '78, p. 255.] Theoretically (see § 565), 
the thickness of the walls of the tube at the lower end being 
considerable, we have 

o=Vgh' ......... . (1) 

as a relation between c, the velocity of the particles impinging 
on the lower end, and the static height h! (§ 565). Eq. (1) is 
verified fairly well by Weisbach's experiments with fine in- 
struments, used with velocities of from 0.32 to 1.24 meters 
per second. Weisbach found 

c = 3.54 Vhf (in meters) met. per sec, 

whereas eq. (1) gives 

c— 3.133 Vhf (in meters) met. per sec. 

In the instruments used by Weisbach the end of the tube 
turning up-stream was probably straight ; i.e., neither flaring 
nor conically convergent. A change in this respect alters the 
relation between c and A'; see § 565 for Pitot's and Darcy's 
results. 

Pitot's Tube, though simple, is not so accurate as the ta- 
chometer. 



A 


c 


B 


B \<y 


_ — ~\ 


- i 


\ 




tzv— -^ 




■ — 


gl *V 



CTJEEENT-METEES. 753 

The Hydro-metric Pendulum, a rather uncertain instrument, 
is readily understood from Fig. 606. The side AB, of the 
quadrant ABC, being held vertical, the 
plane of the quadrant is made parallel to 
the current. The angle 6 between the 
cord and the vertical depends on G, the 
effective weight (i.e., actual weight dimin- 
ished by the buoyant effort) of the ball 
(heavier than water), and the amount of P, 
the impulse or horizontal pressure of the 
current against the latter, since the cord 
will take the direction of the resultant R, FlG " 606 - 

for equilibrium. 

Now P (see § 572) for a ball of given size and character of 
surface varies (nearly) as the square of the velocity ; i.e., if P' 
is the impulse on a given stationary ball, when the velocity of 
the current = c' , then for any other velocity c we have 

P' 

P = impulse = — - c* (2) 

C 
P 

From this and the relation tan = -=- we derive 

(t 

c = y/^.Vu^e. (3) 

With a given instrument and a specified system of units, the 
numerical value of the first radical may be determined as a 
single quantity, by experimenting with a known velocity and 
the value of 6 then indicated; and may then, as a constant fac- 
tor, be employed in (3) for finding the value of c for any ob- 
served value of 6 ; but the same units must be used as before- 

540. Velocities in Different Parts of a Transverse Section. — 
The results of velocity-measurements made by many experi- 
menters do not agree in supporting any very definite relation 
between the greatest surface velocity (c Q max- ) of a transverse 



754 MECHANICS OF EtfGHSTEEKING. 

section and the velocities at other points of the section, bnt 
establish a few general propositions : 

1st. In any vertical line the velocity is a maximum quite 
near the surface, and diminishes from that point both toward 
the bottom and toward the surface. 

2d. In any transverse horizontal line the velocity is a maxi- 
mum near the middle of the stream, diminishing toward the 
banks. 

3d. The mean velocity = v, of the whole transverse section, 
i.e., the velocity which must be multiplied by the area, F, of 
the section, to obtain the volume delivered per unit of time, 

Q = Fo, (1) 

is about 83 per cent of the maximum surface velocity (<? max.) 
observed when the air is still ; i.e., 

V = 0.83 X (C. max.) (2) 

Of eight experimenters cited by Prof. Bowser, only one gives 
a value (= 0.62) differing more than .05 from .83, while others 
obtained the values .82, .78, .82, .80, .82, .83. 

In the survey of the Mississippi River by Humphreys and 
Abbot, 1861, it was found that the law of variation of the 
velocity in any given vertical line could be fairly well repre- 
sented by the ordinates of a parabola (Fig. 607) with its axis 
horizontal and its vertex at a distance d x 
= below the surface according to the follow- 
^= ing relation, f" being a number dependent 
on the force of the wind (from for no 
wind to 10 for a hurricane) \ 

^ = [0.317±0.06/"]tf; . . (3) 

where d is the total depth, and the double 
sign is to be taken -|- for an up-stream, — 
fig. 607. for a down-stream, wind. The following 

relations were also based on the results of the survey : 




(putting, for brevity, B = 1.69 -=- Vd + 1.5,) . (4) 



and 



VELOCITIES IN OPEN CHANNEL. 755 

c = c dl -V^v(^)\ (5) 

c n = ic dl + ic d + ^(ic 9 — ic d ), .... (6) 

c H = c m +^vm. (7) 



{These equations are not of homogeneous form, but call for 
the foot and second as units.) 
In (4), (5), and (6), 

c = velocity at any depth z below the surface ; 
c m = mean velocity in the vertical curve ; 

C dl = U12LX. " " " " 

c^ dl == " at mid-depth ; 

c d = velocity at bottom ; 
v = mean velocity of the whole transverse section. 

It was also found that the parameter of the parabola varied 
inversely as the square root of the mean velocity c m of curve. 

In general the bottom velocity (c d ) is somewhat more than J 
the maximum velocity (e dl ) in the same vertical. In the Mis- 
sissippi the velocity at mid-depth in any vertical was found to 
be very nearly .96 of the surface velocity in the same vertical ; 
this fact is important, as it simplifies the approximate gauging 
of a stream. 

541. Gauging a Stream or River. — "Where the relation (eq. (2), 
§ 540) v — .83 (e max .) is not considered accurate enough for 
substitution in Q = Fv to obtain the volume of discharge (or 
delivery) Q of a stream per time-unit, the transverse section 
may be divided into a number of subdivisions as in Fig. 608, 

of widths a x , a„ etc., and ^____^ v t ^__ y 

mean depths d x , d,, etc., ^"^2" V A .^ f"! » ^ 
and the respective mean ' ^^k^i] fe* JS 8 \J^^^ 

velocities, c x , <? 3 , etc., com- '^^^^mm^m^^^ ' 

puted from measurements fig. 608. 

with current-meters ; whence we may write 

Q = a x d x c x + «„<?„<?„ + a z d. A c % -f- etc. ... (7) 



756 



MECHANICS OF ENGINEEKING. 



With a small stream or ditch, however, we may erect a ver- 
tical boarding, and allow 
the water to flow through a 
rectangular notch or over- 
fall, Fig. 609, and after the 
head surface has become 
permanent, measure \ 
(depth of sill below the 
level surface somewhat 
back of boards), and h 
(width) and use the formu- 
lae of § 504; see examples 




Fig. 609. 



in that article. 



542. Uniform Motion in an Open Channel. — We shall now 
consider a straight stream of indefinite length in which the 
flow is steady, i.e., a state of permanency exists, as distin- 
guished from a freshet or a wave. That is, the flow is steady 
when the water assumes fixed values of mean velocity v, and 
sectional area F, on passing a given point of the bed or bank ; 
and the 

Ea. of continuity . . Q = Fv = F v = F 1 v l = constant . . (1). 

holds good whether those sections are equal or not. 

By uniform motion is meant that (the section of the bed 
and banks being of constant size and shape) the slope of the 
bed, the quantity of water (volume = Q) flowing per time- 
unit, and the extent of the wetted perimeter, are so adjusted 
to each other that the mean velocity of flow is the same in all 
transverse sections, and consequently the area and shape of the 
. transverse section is the same at all points ; and the slope of 
the surface = that of the bed. We may therefore consider, 
for simplicity, that we have to deal with a prism of water of 
indefinite length sliding down an inclined rough bed of con- 
stant slope and moving with uniform velocity (viz., the mean 
velocity v common to all the sections) ; that is, there is no ac- 
celeration. Let Fig. 610 show, free, a portion of this prism, 
of length = Z, and having its bases ~| to the bed and surface. 



UNIFOKM MOTION. OPEN CHANNEL. 757 

The hydrostatic pressures at the two ends balance each other 
from the identity of conditions. The only other forces having 




Fig. 610. 

components parallel to the bed and surface are the weight 
G = Fly of the prism (where y = heaviness of water) making 
an angle = s (= slope) with a normal to the surface, and the 
friction between the water and the bed which is parallel to the 
surface. The amount of this friction for the prism in question 
may be expressed as in § 510, viz.: 

P=fric:LfSy^=fwly?-, ... (2) 

in which S = wl = rubbing surface (area) = wetted perimeter, 
w, X length (see § 538), and f an abstract number. Since the 
mass of water in Fig. 610 is supposed to be in relative equili- 
brium, we may apply to it the laws of motion of a rigid body, 
and since the motion is a uniform translation (§ 109) the com- 
ponents, parallel to the surface, of all the forces must balance. 

7 2 

.*. G sin s must = P =fric. ; .*. Fly - =fwly — ; 

if 

whence 

h-f wl v * (W 

or 

; 4 i < 8 > 

in which F -f- w is called i?, the hydraulic mean depth, or 
hydraulic radius. (3) is sometimes expressed by saying that 



758 



MECHANICS OF ENGINEEKING. 



the whole fall, or head, h, is (in uniform motion) absorbed in 
friction-head. Also, since the slope s = h-=r I, we have 




2<7 



/ 



Vfis; or, v = AVBs, 



(4) 



which is of the same form as Chezy's formula in § 519 for a 
very long straight pipe (the slope s of the actual surface in this 
case corresponding to the slope along piezometer-summits in 
that of a closed pipe). In (4) the coefficient A = VZg -±f is 
not, likey, an abstract number, but its numerical value depends 
on the system of units employed. 

542a. Experiments on the Flow of Water in Open Channels. — 
Those of Darcy and Bazin, begun in 1855 and published in 
1865 (" Kecherches Hydrauliques"), were very carefully con- 
ducted with open conduits of a variety of shapes, sizes, slopes, 
and character of surface. In most of these a uniform flow was 
secured before the taking of measurements. The velocities 
ranged between from about 0.5 to 8 or 10 ft. per second, the 
hydraulic radii from 0.03 to 3.0 ft., with deliveries as high as 
182 cub. ft. per second. For example, the following results 
were obtained in the canals of Marseilles and Craponne, the 
quantity A being for the foot and second. The sections were 
nearly all rectangular. See eq. (4) above. 



No. 


(cub. ft. 


R. 


s. 
abs. 


V. 

(ft. per 


A. 
(foot and 


Character of the masonry 




per sec.) 


(ft.) 


numb. 


sec.) 


sec.) 




1 


182.73 


1.504 


.0037 


10.26 


137.1 


Very smooth. 


2 


143.74 


1.774 


.00084 


5.55 


125. 


Quite " 


3 


43.93 


.708 


.029 


11.23 


78.4 




4 


43.93 


.615 


.060 


13.93 


72.5 


Hammered stone. 


5 


43.93 


.881 


.0121 


7.58 


73.5 


Rather rough. 


6 


43.93 


.835 


.014 


8.36 


77.3 




7 


167.68 


2.871 


. 00043 


2.54 


72.2 


Mud and vegetation. 



[In Experiment No. 7 the flow had not fully reached a state 
of permanency.] 

Fteley and Stearns's experiments on the Sudbury conduit at 
Boston, Mass. (Trans. A.S. C. K, '83), from 1878 to 1880, are 
also valuable. This open channel was of brick masonry with 



UNIFOKM MOTION - . BETTER'S FOKMTTLA. 759 

good mortar joints, and about 9 ft. wide ; the depths of water 
ranging from 1.5 to 4.5 ft With plaster of pure cement on 
the bed in one of the experiments the high value of A — 153.6 
was reached (foot and second), with v = 2.805 ft. per second, 
R = 2.111 ft., ,9 = .0001580, and Q = 87.17 cu. ft. per second. 

Captain Cunningham, in his experiments on the Ganges 
Canal at Roorkee, India, in 1881, found A to range from 48 
to 130 (foot and second). 

Humphreys and Abbot's experiments on the Mississippi 
River and branches (see § 540), with values of R = from 2 or 
3 ft. to 72 ft., furnish values of A — from 53 to 167 (foot and 
second). 

542b. Kutter's Formula. — The experiments upon which 
Weisbach based his deductions for/*, the coefficient of fluid 
friction, were scanty and on too small a scale to warrant gener- 
al conclusions. That author considered thaty depended only 
on the velocity, disregarding altogether the degree of rough- 
ness of the bed, and gave a table of values in accordance with 
that view, these values ranging from .0075 for 15 ft. per sec. 
to .0109 for 0.4 ft. per sec. ; but in 1869 Messrs. Kutter and 
Ganguillet, having a much wider range of experimental data 
at command, including those of Darcy and Bazin, and those 
obtained on the Mississippi River, evolved a formula, known 
as Kutter } s Formula, for the uniform motion of water in open 
channels, which is claimed to harmonize in a fairly satisfactory 
manner the chief results of the best experiments in that direc- 
tion. They make the coefficient A in eq. (4) (or rather the 

factor — — contained in A) a function of R, s, and also n an 

abstract number, or coefficient of roughness, depending on the 
nature of the surface of the bed and banks ; viz., 



v in 

ft. 

per 

sec. 



41.6 + 1^1 + ^81 

n 



-r— -0028 T\ n VB(mft.)xs, . . (5) 

-4- 141,0 4- — — I 

v * V^(infeet); 



which is Kutter's Formula. 



760 MECHANICS OF ENGINEEKING. 

That is, comparing (5) with (4), we have f a function of n, 
R, and s. as follows : 



/= 



+ [41-6 - 



.002811 n 



s J \/R in ft. 



n s 



(6) 



From (6) it appears that f decreases with an increasing R 9 
as has been also noted in the case of closed pipes (§ 517) ; that 
it increases with increasing roughness of surface ; and that it 
is somewhat dependent on the slope. The makers of the 
formula give the following values for n. 

Values of n. n = 

,009 for well-planed timber bed ; 

.010 for plaster in pure cement ; 

.011 for plaster in cement with ^ sand ; 

.012 for unplaned timber ; 

.013 for ashlar and brickwork ; * 

.015 for canvas lining on frames ; 

.017 for rubble ; 

.020 for canals in very firm gravel ; 

.025 for rivers and canals in perfect order and regimen, and 
perfectly free from stones and weeds ; 

.030 for rivers and canals in moderately good order and regi- 
men, having stones and weeds occasionally ; 

.035 for rivers and canals in bad order and regimen, overgrown 
with vegetation and strewn with stones or detritus of 
any sort. 

Kutter's Formula is claimed to apply to all kinds and sizes of 
watercourses, from large rivers to sewers and ditches ; for uni- 
form motion. If VR is the unknown quantity, Kutter's For- 
mula leads to a quadratic equation ; if s the slope, to a cubic. 
Hence, to save computation, tables have been prepared, some 
of which will be found in vol. 28 of Yan ISTostrand's Magazine 

* For ordinary brick sewers Mr. R. F. Hartford claims that n = .014 
gives good results. See Jour. Eug. Societies for '84- '85, p. 220. 



UNIFORM MOTION. OPEN CHANNEL. 



761 



variation of the coefficient A % 



(pp. 135 and 393) (sewers), and in Jackson's works on Hydrau- 
lics (rivers). 

The following table will give the student an idea of the 

-J-, of eq. (4), or large 

bracket of eq. (5), with different hydraulic radii, slopes, and 
values of ?i, according to Kutter's Formula ; from R = \ ft., 
for a small ditch or sluice-way (or a wide and shallow stream), 
to R = 15 ft, for a river or canal of considerable size. Under 
each value of R are given two values of A ; one for a slope of 
s' = .001, and the other for s" — .00005. All these values of 
A imply the use of the foot and second. 

These values of A have been scaled by the writer from a 
diagram given in Jackson's translation of Kutter's " Hydraulic 
Tables,'' and are therefore only approximate. The corre- 
sponding values of f the coefficient of fluid friction, can be 



computed from f 



_ H 



n 


R = 


*ft. 


E = 


:1ft. 


R = 


:3 ft. 


R = 


:6 ft. 


B = 15 ft. 


for 


for 


for 


for 


for 


for 


for 


for 


for for 




s' 


s" 


s' 


s" 


s' 


s" 


s' 


s" 


s' s" 


0.010 


133 


114 


149 


137 


174 


174 


187 


196 


199 222 


0.015 


83 


68 


96 


87 


118 


118 


128 


137 


138 158 


0.020 


58 


49 


70 


63 


87 


87 


98 


106 


110 126 


0.025 


45 


38 


54 


48 


70 


70 


80 


85 


90 106 


0.030 


36 


31 


43 


40 


58 


58 


66 


72 


77 90 


0.035 


28 


25 


38 


34 


50 


50 


58 


64 


68 81 



The formula used in designing the New Aqueduct for New 
York City, in 1885, by Mr. Fteley, consulting engineer, was 

[see (4)] 

v (ft. per sec.) = 142 VR(inft) X s, ... (7) 

whereas Kutter's Formula gives for the same case {a circular* 
section of 14 ft. diameter, and slope of 0.7 ft. to the mile), 
with n = 0.013, 



(ft. per sec.) = 140.7 VR(inft.) X *. 



(8) 



* The aqueduct has this circular form for a small portion, only, of its 
length; a "horseshoe" section of very nearly the same flowing capacity 
being given to the greater portioD of the remainder. 



J 

762 MECHANICS OF ENGINEEKINGr. 

To quote from a letter of Mr. I. A. Shaler of the Aqueduct 
Corps of Engineers, " Mr. Fteley states that the cleanliness of 
the conduit (Sudbury) had much to do in affecting the flow. 
He found the flow to be increased by 7 or 8 per cent in a por- 
tion which had been washed with a thin wash of Portland 
cement." 

Example 1. — A canal 1000 ft. long of the trapezoidal sec- 
tion in Fig. 611 is required to deliver 300 cubic ft. of water 
, per second with the water 8 ft. deep at all 

sections (i.e., with uniform motion), the 
slope of the bank being such that for a depth 
of 8 ft. the width of the water surface (or 
fig. 611. length of air-profile) will be 20 ft.; and the 

coefficient for roughness being n = .020. What is the neces- 
sary slope to be given to the bed (slope of bed = that of sur- 
face, here) (ft., lb., sec.) % 
The mean velocity 

v = Q -^ F= 300 ■+ i (20 + 8) 8 = 2.67 ft. per sec. 

[So that the surface velocity of mid-channel in any section 
would probably be (<? oma x) = 0-5- 0.83 = 3.21 ft. per sec. (eq. 
(2), § 540).] 

The wetted perimeter 



w = 8 + 2 VS 2 + 6 2 = 28 ft, 

and therefore the mean hydraulic depth 

= R = F+ w '= 112 -f- 28 = 4 ft. 

, To obtain a first approximation for the slope, we may use 
the value/ = .00795 given by Weisbach for a velocity of 2.67 
ft. per sec, and obtain, from (3), 

.00795 X 1000 X 28(2.67) a = gl f . 
112X2X32.2 ' '' 

i.e., s — h -f- 1 =■ .000221. 



UNIFORM MOTION IN OPEN CHANNEL. 



763 



With this value for the slope and B = 4 f t. (see above), we 
then have, from eq. (6) (putting n = .020), 



f = 



1 + 41.6 + 



.00281 \ .020 



.000221 



1 V4: ft. 



MM +=4 



.00035 



020 ' .000221 



.0071, 



with which value of f we now obtain 

h = 0.200 feet : i.e., slope = s = .00020. 

Example 2. — If the bed of a creek falls 20 inches every 
1500 ft. of length, what volume of water must be flowing to 
maintain a uniform mean depth of 4J ft., the corresponding 
surface- width being 40 ft., and wetted perimeter 46 ft. ? The 
bed is " in moderately good order and regimen ;" use Kutter's 
Formula, putting n = 0.030 (ft. and sec). 

First we have 



VEs = W(40 X H) -i- (46 X ^~) = .066, 



while VB (ft.) =1.98, and the slope = s = -f-f -f- 1500=.00111 ; 
hence 



v = 



[ 



41.6 + Mil + £02811 

T .030 ^.00111 J 104.43 X .066 



1 + [41. 



or 



.00281 1 0.030 
.00111 J 1.98 



v = 4.13 ft. per sec. 



1.6685 



Hence, also, 

Q = Fv = 40xHX 4.13 = 743.4 cub. ft. per sec. 

[N.B. Weisbach works this same example by eq. (3) with a 
value of f taken from his own table, his result being v = 6.1 




764 MECHANICS OF ENGINEERING. 

ft. per sec, which would probably be attained in practice only 
by making the bed and banks smoother than as given.] 

Example 3. — The desired transverse water-section of a canal 
is given in Fig. 612. The slope is to be 
3 ft. in 1600 ; i.e., s = 3+ 1600 ; or, for 
I = 1600 ft., h = 3 ft. What must be the 
velocity (mean) of each section, for a uni- 
form motion / the corresponding volume 
delivered per sec, Q, = Fv, == ? ; assuming that the character 
of the surface warrants the value n = .030 ? 

Knowing the slope s, = 3 -=- 1600 ; and the hydraulic radius 
p^ = F^ w, = 79.28 sq. ft. -s- 24.67 ft., = 3.215 feet ; with 
n = .030 we substitute directly in eq. (5), obtaining v = 4.67 
ft. per sec ; whence Q = Fv = 370 cub. ft. per sec. 

543. Hydraulic Mean Depth for a Minimum Frictional Resist- 
ance. — We note, from eq. (3), § 542, that if an open channel 
of given length I and sectional area F is to deliver a given 
volume, Q, per time- unit with uniform motion, so that the 
common mean velocity v of all sections (= Q -=- F) is also a 
given quantity, the necessary fall = A, or slope s — h -^ I, is 
seen to be inversely proportional to R, the hydraulic mean 
depth of the section, = (F-t- w), = sectional area -5- wetted 
perimeter. 

For h to be as small as possible, we may design the form of 
transverse section, so as to make R as large as possible ; i.e., 
to make the wetted perimeter a minimum for a given F\ for 
in this way a minimum of frictional contact, or area of rub- 
bing surface, is obtained for a prism of water of given sectional 
area i^and given length I. 

In a closed pipe running full the wetted perimeter is the 
whole perimeter; and if the given sectional area is shaped in 
the form of a circle, the wetted perimeter, == w, is a minimum 
(and R a maximum). If the full pipe must have a polygonal 
shape of n sides, then the regular polygon of n sides will pro- 
vide a minimum w. 

Whence it follows that if the pipe or channel is running 



UNIFOKM MOTION IN OPEN CHANNEL. 



765 




Fig. 613. 



half full, and thus becomes an open channel, the semicircle, 

of all curvilinear water pro- 
files, gives a minimum w. 
Also, of all trapezoidal pro- 
files with banks at 60° with 
the horizontal the half of a 
regular hexagon gives a 
minimum w. Among all 
rectangular sections the half 
square gives a minimum w ; 
and of all half octagons the half of a regular octagon gives a 
minimum w (and max. E) for a given F. See Fig. 613 for 
all these. 

The egg-shaped outline, Fig. 614, small end down, is fre- 
quently given to sewers in which it is important that the 
different velocities of the water at dif- 
ferent stages (depths) of flow (depend- 
ing on the volume of liquid passing per 
unit-time) should not vary widely from 
each other. The lower portion ABC, 
providing for the lowest stage of flow 
AB, is nearly semicircular, and thus in- 
duces a velocity of flow (the slope being 
constant at all stages) which does not 
differ extremely from that occurring 
when the water flows at its highest 
stage DE, although this latter velocity is the greater; the 
reason being that ABC from its advantageous form has a 
hydraulic radius, R, larger in proportion to its sectional area, 
F, than DCE. 

That is, F -$-w for ABC is more nearly equal to F-i- w for 
DEC than if DEC were a semicircle, and the velocity at the 
lowest stage may still be sufficiently great to prevent the de- 
posit of sediment. See § 575. 

544. Trapezoid of Fixed Side-slope. — For large artificial water- 
courses and canals the trapezoid, or three-sided water-profile 
(symmetrical), is customary, and the inclination of the bank, 




Fig. 614. 



766 MECHANICS OF ENGINEEKING. 

or angle 8 with the horizontal, Fig. 615, is often determined 

Iw /^ ^y the nature of the material 

^ I fc r- ^^ 7-^ | " ^p ^^ y composing it, to guard against 

p^~=\^*x [T ^^ washouts, caving in, etc. We 

j j||i~iEErv > ■! are therefore concerned with the 

WMS^M?®^^ F following problem : Given the 
FlG - 615 - area, F, of the transverse section, 

and the angle 8, required the value of the depth x (or of upper 
width z, or of lower width y, both of which are functions of x) 
to make the hydraulic mean depth, B = F ' -\-w,a maximum, 
or w -f- F a minimum. F is constant. 
From the figure we have 



and 
whence 



w = AB + 2BC = y + 2x cosec. 8, . . . (1) 
F '= yx -f- x* cot. 8; 

y = -.(F-x' 1 cot.8), (2) 

x 



substituting which in (1) and dividing by F, noting that 

2 cosec. 8 — cot. 8 = ; — - — , we have 

sin 8 

w 1 1 . 2 — cos 8 /ox 

F=B = x+-F^T- X (3) 

For a minimum w we put 

d \FJ „ . 1 . 2-cos0 



= 0; i.e., ..-^—^Q; 



x (tor max. or mm. w) = ± a / - 



cos 



The -)- sign makes the second derivative positive, and hence 
for a min. w or max. R we have 



a? (call it a/) — x f — — ■■, ... (4) 

V ' V2 - cos 8 ' V ' 



TRAPEZOID FOE MINIMUM WETTED PERIMETER. 767 

while the corresponding values for the other dimensions are 

y' =.^--x' cot. (5) 

x 

and 

z f =y' + 2x'cot. 6 = ^- + x'qoL 6. ... (6) 
x 

For the corresponding hydraulic mean depth R' [see (3)], 
i.e., the max. R, we have 

1 1 2 — cos e , _ _2_ _ 



„, i , 1 / i^sinfl 



(8) 



Equations (4), (5), ... (8) hold good, then, for the trapezoi- 
dal section of least frictional resistance for a given angle 6. 

Pkoblem. — Required the dimensions of the trapezoidal sec- 
tion of minimum frictional resistance for = 45°, which with 
h = 6 inches fall in every 1200 feet (= I) is required to de- 
liver Q = 360 cub. ft. of water per minute with uniform 
motion. 

Here we have given, with uniform motion, A, Z, and Q 9 
with the requirement that the section shall be trapezoidal, with 
6 = 45°, and of minimum frictional resistance. The following 
equations are available : 

Eq. of continuity . . . Q = ff% (!') 



Eq. (8) preceding, for con- ) jgr = \ I sing i /jr ^ ,%) 
dition of least resistance J " 2 y 2 — cos 6 ' 

fs)^!; } foruniformmotion, A=y^.|- <*) 

There are three unknown quantities, v, F, and R'. Solve 



768 



MECHANICS OF ENGINEERING. 



(1') for v; solve (2') fori?'; substitute their values in (3'); 
whence 



hVsin 6 __ f_ 
4/2-cos(9~ ty 



21 



VF F 






2gh Vsin 6 



Since f cannot be exactly computed in advance, for want of 
knowing the value of R, we calculate it approximately [eq. (6)v 
§ 542b] for an assumed value of i?, insert it in the above 
equation (4'), and thus find an approximate value of F; and 
then, from (8), a corresponding value of R, from which a new 
value of yean be computed. Thus after one or two trials a 
satisfactory adjustment of dimensions can be secured. 

545. Variable Motion. — If a steady flow of water of a de- 
livery Q, = Fv, = constant, takes place in a straight open 
channel the slope of whose bed has not the proper value to 
maintain a " uniform motion" then " variable motion? ensues 
(the flow is still steady, however); i.e., although the mean 
velocity in any one transverse section remains fixed (with lapse 
of time), this velocity has different values for different sections ; 
but as the eq. of continuity, 

Q = Fv = F 1 v 1 = Ff> 1 , etc., 

still holds (since the flow is steady), the different sections 

have different areas. If, 
Fig. 616, a stream of 
water flows down an 
inclined trough without 
friction, the relation 
between the velocities 
v and v l at any two 
sections and 1 will be 

the same as for a material point sliding down a guide without 

friction (see § 79, latter part), viz. : 




Fig. 616. 



V* 



2a 



+ h, 



(I) 



VARIABLE MOTION. OPEN CHANNEL. 769 

an equation of beads (really a case of Bernoulli's Theorem, 
§ 492). But, considering friction on the bed, we must sub- 

7 2 

tract the mean friction-head f ' -^ . -^ [see eqs. (3) and (3'), 
§ 542] lost between and 1 ; this friction-head may also be 
written thus: / '-— -™-^- : and therefore eq. (1) becomes 

*-* + *-^'5P •■•.•• (2) 

which is the formula for variable motion j and in it I is the 
length of the section considered, which should be taken short 
enough to consider the surface straight between the end-sec- 
tions, and the latter should differ but slightly in area. The 
subscript m may be taken as referring to the section midway 
between the ends, so that v m * = i(v* -f- v^). The wetted pe- 
rimeter w m = %(w + w,), and F m — i(F + F x ). Hence eq. 
(2) becomes 

h ~2i~^ + ~^+F l 2g—' * ' (> 

and again, by putting v = Q -~ F , v l = Q -r- F 1 , we may 
write 



rj_ 1 1 fl{w a + w s ) ll 1 n 



2<7 



whence 



<g = , ^ (5) 

/i ii /zk + ^) r l in w 
Vi^ 2 J p;«" r 2" i^ + i^ 'L^ 7 2 ' r ^ 2 J 

From eq. (4), having given the desired shapes, areas, etc., of 
the end-sections and the volume of water, Q, to be carried per 
unit of time, we may compute the necessary fall, A, of the sur- 
face, in length = I ; while from eq. (5), having observed in an 
actual water-course the values of the sectional areas F Q and F x . 
the wetted perimeters w and w 1 , the length, = l 3 of the por 



770 



MECHANICS OF EJN GXNEEEING. 



tion considered, we may calculate Q and thus gauge the stream 
approximately, without making any velocity measurements. 

As to the value of f, we compute it from eq. (6), § 542b, 
using for R a mean between the values of the hydraulic radii 
of the end-sections. 

546. Bends in an Open Channel. — According to Humphreys 
and Abbot's researches on the Mississippi River the loss of 
head due to a bend may be put 



K = 



536 n 



(1) 



in which v must be in ft. per sec, and tf, the angle ABC, Fig. 
617, must be in 7r-measure, i.e. in radians. 
The section F must be greater than 100 
sq. ft., and the slope s less than .0008. v 
is the mean velocity of the water. Hence 
if a bend occurred in a portion of a 
stream of length I, eq. (3) of § 542 be- 
comes 




Fig. 617. 



B 2g 



6_^# 
536 n 



while eq. (2) of § 545 for variable motion would then become 

V A = < +A_^<_ -*?-*. . (ft. and sec). . (3) 

2g 2g^ F m 2g 536 n * > w 

(v and S as above.) (For "radian" see p. 544.) 

547. Equations for Variable Motion, introducing the Depths. 
—Fig. 618. The slope of the bed being sin a (or simply a, 
Trmeas.), while that of the surface is 
different, viz., ^^^-_----^-~-^ ^ 

sin /3 = s = h -r- 1, 
we may write 

h = d Q + I sin a — d. , 

FIG.61& 




YAEIABLE MOTION. OPEN CHANNEL. 771 

in which d and d x are the depths at the end-sections of the 
portion considered (steady flow with variable motion). "With 
these substitutions in eq. (4), § 545, we have, solving for I, 

i = ^ — ° • g — . . . . a) 

F+fM+f^ sma 

From which, knowing the slope of the bed and the shape 
and size of the end-sections, also the discharge Q, we may 
compute the length or distance, I, between two sections whose 
depths differ by an assigned amount (d — d x ). But we can- 
not compute the change of depth for an assigned length I from 
(6). However, if the width b of the stream is constant, and 
the same at all depths ; i.e., if all sections are rectangles hav- 
ing a common width ; eq. (6) may be much simplified by intro- 
ducing some approximations, as follows : We may put 

\f; F:h g f:f: > f: > 

_ \_o 1/\ i — iy o W ] 11C ] 1 approx. = -^ JL r= — L/ • ~ v 

and, similarly, 



"Sf:^ f?) 



Q> = Wm (F; + F 1 > v^ 

f + f x \f: ' *;•; 2^ (f + f,) f: ' % g 

which approx. = —^ — - . 
Hence by substitution in eq. (6) we have 



Z = 






(7) 



547a. Backwater. — Let us suppose that a steady flow has 
been proceeding with uniform motion (i.e., the surface parallel 



772 



MECHANICS OF ENGINEERING. 



to the bed) in an open channel of indefinite extent, and that a 
vertical wall is now set up across the stream. The water rises 
and flows over the edge of the wall, or weir, and after a time 
a steady flow is again established. The depth, y , of the water 
close to the weir on the up-stream side is greater than d , the 
original depth. We now have " variable motion " above the 
weir, and at any distance x up-stream from the weir the new 
depth y is greater than d . This increase of depth is called 
backwater, and, though decreasing up-stream, may be percep- 
tible several miles above the weir. Let s be the slope of the 
original uniform motion (and also of present bed), and v the 



velocity of the original uniform motion, and let k = 



V 



Then, if the section of the stream is a shallow rectangle of 
constant width, we have the following relation (Rankine) : 



x = 



y -y + (d -2k)(cp-cp ) 



y 



(1) 



where is a function of ^-, as per following table : 



For|- = 1.0 
do 

= OO 


1.10 
.680 


1.20 
.480 


1.30 
.376 


1.40 
.304 


1.50 
.255 


1.60 
.218 


1.70 
.189 


For|- =1.80 
do 
0= .166 


1.90 
.147 


2.00 
.132 


2.20 

.107 


2.40 
.089 


2.60 
.076 


2.80 
.065 


3.0 
.056 



O is found from ^-, precisely as from ^-, by use of the table. 

With this table and eq. (1), therefore, we can find a?, the dis- 
tance (" amplitude of backwater") from the weir of the point 
where any assigned depth y (or " height of backwater," y — d ) 
will be found. 

For example, Prof. Bowser cites the case from D'Aubuis- 
son's Hydraulics of the river Weser in Germany, where the 
erection of a weir increased the depth at the weir from 2.5 ft. 
to 10 ft., the flow having been originally " uniform" for 10 
miles. Three miles above the dam the increase (y — d ) of 
depth was 1.25 ft., and even at four miles it was 0.75 ft. 



CHAPTER VIII. 

DYNAMICS OF GASEOUS FLUIDS. 

548. Steady Flow of a Gas. — [KB. The student should no^r 
review § 492 up to eq. (5).] The differential equation from 
which Bernoulli's Theorem was derived for any liquid, with- 
out friction, was [eq. (5), § 492] 

- vdv -\- dz -)- - dp = 0, (A) 

if / 

and is equally applicable to the steady flow of a gaseous fluid, 
but with this difference in subsequent work, that the heaviness, 
Y (§ <0> °f the g as passing different sections of the pipe or 
stream-line is, or may be, different (though always the same at 
a given point or section, since the flow is steady). For the 
present we neglect friction and consider the flow from a large 
receiver, where the great body of the gas is practically at rest, 
through an orifice in a thin plate, or a short nozzle with a 
rounded entrance. 

In the steady flow of a gas, since y is different at different 
points, the equation of continuity takes the form 

Flow of weightier time-unit = F 1 v 1 y 1 = F^v 2 y^ = etc. ; . (a) 

i.e., the weight of gas passing any section, of area F, per unit 
of time, is the same as for any other section, or Fvy = con- 
stant, y being the heaviness at the section, and v the velocity. 

549. Flow through an Orifice — Remarks. — In Fig. 619 we 
have a large rigid receiver containing gas at some tension,^, 
higher than that, p m , of the (still) outside air (or gas), and at 
some absolute temperature T n , and of some heaviness y n \ that 
is, in a state n. The small orifice of area F being opened, the 
gas begins to escape, and if the receiver is very large, or if the 
supply is continually kept up (by a blowing-engine, e.g.), after 

773 



774 



MECHANICS OF ENGINEEKHSTG. 



£ 






X\^ 



8TILL. 
. AIR- . 



£W-:^ 



|§L*. 






IP..HT 



Fig. 619. 



a very short time the flow becomes steady. Let nm represent 
any stream-line (§ 495) of the flow. According to the ideal 
subdivision of this stream-line into 
laminae of equal mass or weight (not 
equal volume, necessarily) in estab- 
lishing eq. (A) for any one lamina, 
each lamina in the lapse of time dt 
moves into the position just vacated 
by the lamina next in front, and 
assumes precisely the same velocity, 
pressure, and volume {and there- 
fore heaviness) as that front one had at the beginning of the 
dt. In its progress toward the orifice it expands in volume, 
its tension diminishes, while its velocity, insensible at n, is 
gradually accelerated on account of the pressure from behind 
always being greater than that in front, until at m, in the 
" throat" of the jet, the velocity has become v m , the pressure 
(i.e., tension) has fallen to a value p m , and the heaviness has 
changed to y m . The temperature ^(absolute) is less than 
T n , since the expansion has been rapid, and does not depend 
on the temperature of the outside air or gas into which efflux 
takes place, though, of course, after the effluent gas is once 
free from the orifice it may change its temperature in time. 

We assume the pressure p m (in throat of jet) to be equal to 
that of the outside medium (as was done with flow of water), 
so long as that outside tension is greater than .527 j? n ; but if it 
is less than .527 p n and is even zero (a vacuum), experiment 
seems to show that p m remains equal to 0.527 of the interior 
tension p n : probably on account of the expansion of the 
effluent gas beyond the throat, Fig. 620, so 
that although the tension in the outer edge, 
at a, of the jet is equal to that of the outside 
medium, the tension at m is greater because 
of the centripetal and centrifugal forces devel- 
fig. 620. oped in the curved filaments between a and 

m. (See § 553.) 

550. Flow through an Orifice; Heaviness assumed Constant 
during Flow. The Water Formula. — If the inner tension p n ex- 



f 



-& 

'/><* 



^ 



STEADY FLOW OF GASES. 775 

ceeds the outer, p m , but slightly, we may assume that, like 
water, the gas remains of the same heaviness during flow. 
Then, for the simultaneous advance made by all the laminae of 
a stream-line, Fig. 619, in the time dt, we may conceive an 
equation like eq. (A) written out for each lamina between n 
and m } and corresponding terms added ; i.e., 

{For orifice*) .... Ifjdv + fjz + £"f = 0. . (i?) 

In general, y is different in the different laminae, but in the 
present case it is assumed to be the same in all ; hence, -with 
m as datum level and h = vertical distance from n to m, we 
have, from eq. (B), 

^ - ^L. -i- - h + S^ — £± = 0. . . . (1) 
2g 2g ^ ^ y y W 

But we may put v n = ; while h, even if several feet, is 
small compared with — — — . E.g., with p m = 15 lbs. per 

sq. in. and p n = 16 lbs. per sq. in., we have for atmospheric 
air at freezing temperature 

y r 

Hence, putting v n = and h = in eq. (1), we have 

v m __ JPn —Pm j Water formula • for small \ /o\ 

2g y n \ difference of pressures, only. ) ^ ' 

The interior absolute temperature T n being known, the y n 
(interior heaviness) may be obtained from y n =p n y T ~ T n p 
(§ 472), and the volume of flow per unit of time then obtained 
(first solving (2) for v m ) is 

Q m = F m v m9 (3) 

where F m is the sectional area of the jet at m. If the mouth- 
piece or orifice has well-rounded interior edges, as in Fig. 541, 



776 MECHANICS OF ENGINEERING. 

its sectional area F may be taken as the area F m . But if it is 
an orifice in " thin plate," putting the coefficient of contraction 
= C= 0.60, we have 

F m =CF = 0.60 F; and Q m = 0.60 Fv m . . (4) 

This volume, Q m , is that occupied by the flow per time-unit 
when in state m, and we have assumed that y m = y n ; hence 
the weight of flow per time- unit is 

& = QmYm = ■^m^mrm = F m V m y n . . /-. . (5) 

Example. — In the testing of a blowing-engine it is found 
capable of maintaining a pressure of 18 lbs. per sq. inch in a 
large receiver, from whose side a blast is steadily escaping 
through a " thin plate" orifice (circular) having an area F = 4 
sq. inches. The interior temperature is 30° Cent, and the out- 
side tension 15 lbs. per sq. in. 

Required the discharge of air per second, both volume and 
weight. The data are : jp n = 18 lbs. per sq. in., T n = 303° 
Abs. Cent., F— 4 sq. inches, and p m = 15 lbs. per sq. in. Use 
ft.-lb.-sec. system. 

First, the heaviness in the receiver is 

^=t%° = ^-S x - 0807 =- 089 lbs - * er cub - ft 

Then, from eq. (2), 



/ZEEK- / 2X32.2[144X 18-144x15] 

W= V ~^~ v — ™ ~ 



555.3 

feet 
per sec. 



(97 per cent of this would be more correct on account of fric- 
tion.) 

.-. Q m =F m v m =.6Fv m = ^.^xteM = 9.24 cub. ft. per sec. 

at a tension of 15 lbs. per sq. in., and of heaviness (by hypoth- 
esis) = .089 lbs. per cub. ft. Hence weight 

= Q = 9.24 X .089 = .82 lbs. per sec. 



FLOW OF GASES BY MATCIOTTE'S LAW. 777 

The theoretical power of the air-compressor or blowing-en- 
gine to maintain this steady flow can be computed as in Exam- 
ple 3, § 483. 

551. Flow through au Orifice on the Basis of Mariotte's Law ; 
or Isothermal Efflux. — Since in reality the gas expands during 
flow through an orifice, and hence changes its heaviness (Fig. 
619), we approximate more nearly to the truth in assuming 
this change of density to follow Mariotte's law, i.e., that the 
heaviness varies directly as the pressure, and thus imply that 
the temperature remains unchanged during the flow. We 
again integrate the terms of eq. (B\ but take care to note that, 
now, y is variable (i.e., different in different laminse at the 
same instant), and hence express it in terms of the variable p 
(from eq. (2), § 475), thus : 

y = {y n +Pn)p- 
Therefore the term / -^- of eq. (B) becomes 

t/n y 
Yn Jn P Yn *'W VJ 

and, integrating all the terms of eq. (Z?), neglecting A, and call- 
ing v n zero, we have 

v _™_ _ Pn j Pn j efflux hy Mariotte's ) ,%) 

ty Yn Pm ' \ Law through orifice j ' ' W 

T v 

As before, y n — -^ f . — y Q , and the flow of volume per time- 

■*- n Po 

unit at m is 

Q m = F m v m ; (3) 

while if the orifice is in thin plate, F m may be put = .60 F , 
and the 



wet 



ght of the flow per time-unit = G — F m v m y m . . (4) 



If the mouth-piece is rounded, F m — F= area of exit orifice 
of mouth-piece. 



778 MECHANICS OF ENGINEEKING. 

Example. — Applying eq. (2) to the data of the example in 
§ 550, where y n was found to be .089 lbs. per cub. ft., we have 
[ft., lb., sec] 



V Yn Pm 



.*• Qm = F m v m = 0.60 X xt* X 584.7 = 9.745 cub. ft. per sec. 
Since the heaviness at m is, from Mariotte's law, 

y m = ^ y n = f| of .089, i.e., y m = .0741 lbs. per cub. ft., 

hence the weight of the discharge is 

G = QmYm = 9.745 X .0741 = 0.722 lbs. per sec, 

or about 12 per cent less than that given by the " water for- 
mula." If the difference between the inner and outer tensions 
had been less, the discrepancy between the results of the two 
methods would not have been so marked. 

552. Adiabatic Efflux from an Orifice. — It is most logical to 
assume that the expansion of the gas approaching the orifice, 
being rapid, is adiabatic (§ 478). Hence (especially when the 
difference between the inner and outer tensions is considerable) 
it is more accurate to assume y as varying according to Pois- 
son's Law, eq. (1), § 478 ; i.e., y — [y n -±pj]p% in integrat- 
ing eq. (B). Then the term 



Y Y* 



--^[•-(SJll 



ADIABATIC FLOW OF GASES THROUGH ORIFICES. 779 

and eq. (B), neglecting h as before, and with v n = 0, becomes 
(See Fig. 619) 



Yn 






. (Adiabatic flow ; orifice.) . (1) 



Having observed p n and T n in the reservoir, we compute 

rri 

y n = '?™Y±_^ (from § 472). The gas at m, just leaving the 

orifice, having expanded adiabatically from the state n to the 
state m, has cooled to a temperature T m (absolute) found thus 
(§478), 

T m =T n (SA\ (2) 

and is of a heaviness 

v m = yJH % , (3) 

and the flow per second occupies a volume (immediately on 
exit) 

Q m — F m v m , (4) 

and weighs 

G — F m v m y m (5> 

Example 1. — Let the interior conditions in the large reser- 
voir of Fig. 619 be as follows {state n) : p n — 22^- lbs. per sq. 
in., and T n = 294° Abs. Cent, (i.e., 21° Cent); while ex- 
ternally the tension is 15 lbs. per sq. inch, which may be taken 
as being = p m — tension at m, the throat of jet. The opening 
is a circular orifice in " thin plate" and of one inch diameter. 
Kequired the weight of the discharge per second [ft., lb., sec; 
<7 = 32.2]. 

First, Yn = 2 f; 5X / - 4 i^ X .0807 = 0.114 lbs. per cub. ft. 
15 X i44 294 

Then, from (1), 



s/ 



pj .-J 



2 X 32.2 X 3 X 22.5 X 144 n 3 __ 

0^4 t ~~ ^f ] = ft " per sec ' 



780 MECHANICS OP ENGINEEKING. 

JSTow F = \n{^y= .00546 sq. ft. 

.\ Q m = CFv m =.60Fv m = 0.60 X .00546 X 844 = 2.765 
cub. ft. per sec, at a temperature of 

T m = 294 Vf = 257° Abs. Cent. = - 16° Cent., 
and of a heaviness 

Ym = 0.114 V(f7= 0.085 lbs. per cub. ft., 
so that the weight of flow per sec. 

= G= QmYm = 2.765 X .085 = .235 lbs. per sec. 

Example 2. — Let us treat the example already solved by the 
two preceding approximate methods (§§ 550 and 551) by the 
present more accurate equation of adiabatic flow, eq. (1). 

The data were (Fig. 619) : 

p n = 18 lbs. per sq. in. ; T n = 303° Abs. Cent. ; 
p m = 15 " " " ; and F = 4 sq. inches 

[F being the area of orifice]. y n was found = .089 lbs. per 
cub. ft. in § 550 ; hence, from eq. (1), 



V 



2 X 32.2 X 3 X 18 X 144 n * /Tl Kh7R .,. 
Ogg [1— V|]=:576.2ft.persec. 

From (4), 

Q m =F m v m =.6Fv m =.6x T i T X M6.2 = 9.603 cub. ft. per sec; 

and since at m it is of a heaviness 

Vm = .089V(f§y 2 = .0788 lbs. per cub. ft., 
we have weight of flow per sec. 

= G — Q m y m = 9.603 X .0788 = 0.756 lbs. per sec. 



THEOKETICAL MAXIMUM FLOW OF WEIGHT OF GAS. 781 

Comparing the three methods for this problem, we see that 

By the " water formula" . . . G = 0.82 lbs. per sec. 
" isothermal formula, . . G = 0.722 " " 
" adiabatic formula, . . G = 0.756 " " 

553. Practical Notes. Theoretical Maximum Flow of Weight. 

— If in the equations of § 552 we write for brevity p m -t-p n = x 
we derive, by substitution from (1) and (3) in (5), 

nerS. of W \ = G = QmYm=F m V6gp n y n [1-x^xl . (1) 



per unit of time 

This function of x is of such a form as to be a maximum for 

« = (i>»-iO = (i) 3 =.512; .... (2) 

i.e., theoretically, if the state n inside the reservoir remains 
the same, while the outside tension (considered =p m of jet, 
Fig. 619) is made to assume lower and lower values (so that 
x, =£> m -r-pnt diminishes in the same ratio), the maximum flow 
of weight per unit of time will occur when p m = .512 p n , a 
little more than half the inside tension. (With the more ac- 
curate value 1.41 (1.408), instead of f , see § 478, we should 
obtain .527 instead of .512 for dry air; see § 549.) 

Prof. Cotterill says (p. 544 of his " Applied Mechanics") : 
" The diminution of the theoretical discharge on diminution 
of the external pressure below the limit just now given is an 
anomaly which had always been considered as requiring ex- 
planation, and M. St. Yenant had already suggested that it 
could not actually occur. In 1866 Mr. R. D. Napier showed 
by experiment that the weight of steam of given pressure dis- 
charged from an orifice really is independent of the pressure 
of the medium into which efflux takes place * ; and in 1872 
Mr. Wilson confirmed this result by experiments on the reac- 
tion of steam issuing from an orifice." 

" The explanation lies in the fact that the pressure in the 

* When the difference between internal and external pressures is great, — 
should be added. 



732 MECHANICS OF ENGINEERING. 

centre of the contracted jet is not the same as that of the sur- 
rounding medium. The jet after passing the contracted sec- 
tion suddenly expands, and the change of direction of the fluid 
particles gives rise to centrifugal forces" which cause the pres- 
sures to be greater in the centre of the contracted section than 
at the circumference ; see Fig. 620. 

Prof. Cotterill then advises the assumption that j) m =.527p n 
(for air and perfect gases) as the mean tension in the jet at m 
(Fig. 619), whenever the outside medium is at a tension less 
than .527 p n . He also says, "Contraction and friction must 
be allowed for by the use of a coefficient of discharge the 
value of which, however, is more variable than that of the 
corresponding coefficient for an incompressible fluid. Little is 
certainly known on this point." See §§ 549 and 554. 

For air the velocity of this maximum flow of weight is 



Vel of max. G = |~ 997 \ /^"j ft - P er sec -> 



(3) 



where T n = abs. temp, in reservoir, and T = that of freezing 
point. Rankine's Applied Mechanics ( p. 584) mentions ex- 
periments of Drs. Joule and Thomson, in which the circular 
orifices were in a thin plate of copper and of diameters 0.029 
in., 0.053 in., and 0.084 in., while the outside tension was 
about one half of that inside. The results were 84 per cent 
of those demanded by theory, a discrepancy due mainly, as 
Hankine says, to the fact that the actual area of the orifice was 
used in computation instead of the contracted Section; i.e., con- 
traction was neglected. 

554. Coefficients of Efflux by Experiment. For Orifices and 
Short Pipes. Small Difference of Tensions, — Since the discharge 
through an orifice or short pipe from a reservoir is affected 
not only by contraction, but by slight friction at the edges, 
even with a rounded entrance, the theoretical results for the 
volume and weight of flow per unit of time in preceding para- 
graphs should be multiplied both by a coefficient of velocity 
and one for contraction C, as in the case of water ; i.e., by a 
coefficient of efflux /*, = <pC. (Of course, when there is no 



COEFFICIENTS OF EFFLUX. GAS. JB3 

contraction, C= 1.00, and then pi = <p as with a well-rounded 
mouth-piece, for instance. Fig. 541, and with short pipes.) 

Hence for practical results, with orifices and short pipes, we 
should write for the weight of flow per unit of time 



=-* = ^ ra = ^.yW.[i- ffi] • (1) 

(from the equations of § 552 for adiabatic flow, as most accu- 
rate ; p m -^r]) n may range from -J to 1.00). F= area of orifice, 
or of discharging end of mouth-piece or short pipe. y n = 
heaviness of air in reservoir and = T p n y -f- T n p , eq. (13) of 
§ 437 ; and /j. = the experimental coefficient of efflux. 

From his own experiments and those of Koch, B'Aubuis- 
son, and others, "Weisbach recommends the following mean 
values of ju for various mouthpieces, when p n is not more than 
■J- larger than p m (i.e., about 17 $ larger), for use in eq. (1) : 

1. For an orifice in a thin plate, ^=0.56 

2. For ashortcyiindricalpipe(innercornersnotrounded),/*=0.75 

3. For a well-rounded mouth-piece (like that in Fig. 541), /*=0.98 

4. For a short conical convergent pipe (angle about 6°), yu=0.92 

Example. — (Data from Weisbach's Mechanics.) "If the 
sum of the areas of two conical tuyeres of a blowing-machine 
is i^= 3 sq. inches, the temperature in the reservoir 15° Cent., 
the height of the attached (open) mercury manometer (see 
Fig. 464) 3 inches, and the height of the barometer in the ex- 
ternal air 29 inches," we have (ft., lb., sec.) 

f = W^=W ^88°Ab,Cen,;. 
Pn =* (M) 14 ^ X 144 lbs. per sq. ft. ; 
Yn = Ui'U X 0.0807 = 0.0816 lbs. per cub. ft, 
while F= yf^ sq. ft. and (see above) /t = 0.92 ; hence 

G = 0.92 X tH (ft)* ^2 X 32 2 X 3X H X 14.7 X 144 X .0816 [1 - f f|]j 



784 



MECHANICS OF ENGINEERING. 



i.e.. G — .6076 lbs. per second ; which will occupy a volume 

F = G -5- y = G ~ .0807 = 7.59 cab. ft. 

at one atmosphere tension and freezing-point tempoTatare ^ 
while at a temperature of T n = 288° Abs. Cent, and tendon of 
j? m = ||- of one atmosphere (i.e., in the state in which it was 
on entering the blowing-engine) it occupied a volume 

V=¥ti'WX 7-59 = 8.24 cub. ft. 

(This last is Weisbach's result, obtained by an approximate 
formulae) 



555. Coefficients of Efflux for Orifices and Short Pipes for a 
Large Difference of Tension.— -For values > £ and < 2, of the 

ratio p n l p m , of internal to external tension, Weisbach's ex* 
periments with circular orifices in thin plate, of diameters (= d) 
from 0.4 inches to 0.8 inches, gave the following results : 



Pn : Pm = 

." or d = .4?"-; jn = 
" d=.8™-,M = 


1.05 
.55 

.56 


1.09 
.59 

.57 


1.40 
.69 
.64 


1.65 

.72 
.68 


1.90 
.76 


2.00 
.78 

.72 



Whence it appears that /a. increases somewhat with the ratio of 
Pn to Pm > an d decreases slightly for increasing size of orifice. 

With short cylindrical pipes, internal edges not rounded, 
and three times as long as wide, Weisbach obtained p. as 
follows : 



Pn : Pm = 


1.05 


1.10 


1.30 


1.40 


1.70 


1.74 


diam. = .4 in - ; jn = 


.73 


.77 


.83 








" = .6 in -; ^i — 








.81 


.82 




" =1.0 in - ; M = 












,83 



When the inner edges of the 0.4 in. pipe were slightly 
rounded, jjl was found = 0.93 ; while a well-rounded mouth- 
piece of the form shown in Fig. 541 gave a value jj. = from 
,965 to .968, for p n : p m ranging from 1.25 to 2.00, These 
values of /* are for use in eq. (1), above. 

556. To find the Discharge when the Internal Pressure is 
measured in a Small Reservoir or Pipe, not much larger than the 



VELOCITY OF A.PPKOACH. GASES. 785 

Orifice. — Fig. 621. If the internal pressure p n , and tempera- 
ture T n , must be measured in a M II 

small reservoir or pipe, n 9 whose Af pyl It 

sectional area F n is not very large g — =7 ^£3^j^f!_^,^ 

compared with that of the orifice, ■^Jr = =^^ ^^^^^on t 

F, (or of the jet, F m ,) the velocity 3" ^— =^N 

v n at n (velocity of approach) can- fig. 621. 

not be put = zero. Hence, in applying eq. (B), § 550, to the 

successive laminae between n and m, and integrating, we shall 

have, for adiabatic steady flow, 

n^_v^^Zp n r 1 _(Prn\f] m 

%g ty vn L \pj J " ' ' • ■ w 

instead of eq. (1) of § 552. But from the equation of continuity 

for steady flow of gases [eq. (a) of § 548], F n v n y n = F m v m y m ; 

F V 2 
hence v n 2 = J?/™ Vmy while for an adiabatic change from n 

■tn Yn 

to ra, — = \S^\ ; whence by substitution in (1), solving for 
v m> we have 

As before, from §§ 472 and 478, 

Yn = -zr-^- Y» ...... (3) 

and Vm= {fJ Vn ' •••-••• W 

Having observed p n , p m , and T n , tiion, and knowing the 
area F of the orifice, we may compute y n , y m , and v m9 and 
finally the 

Weight of flow per time-unit = G = fiFv m y m ^ . . (5) 



U« = 



(2) 



786 MECHANICS OF ENGINEERING. 

taking /* from § 554 or 555. In eq. (2) it must be remembered 
that for an orifice in " thin plate," F m is the sectional area of 

the contracted vein, and = CF; where C may be put = ~ . 

Example. — If the diameter of AB, Fig. 621, is 3 J inches, 
and that of the orifice, well rounded, = 2 in. ; if p n = l-£% at- 
mospheres = -if X 14.7 X 144 lbs. per sq. ft., while p m = \% of 

an atmos., so that ^ = f§-, and T n = 283° Abs. Cent.,— re- 

quired the discharge per second, using the ft., lb., and sec. 
From eq. (3), 

Tn = U'Ui X 0.0807 <= .08433 lbs. per cub. ft.; 
•whence (eq. (4)) 

Vm = (i|)Vn = .07544 lbs. per cub. ft 
Then, from eq. (2), 

r\/64.4x3xl5.925xl44~ ~~~~] . f ,/= n BWl 1X / 1 

*.= [r :o8433 (i-(H)*)J ^ I ^-(tt) (H) 1 J 

= 558.1 ft. per sec. ; 

.\ G = 0.98 |QV558.1 X .07544 = .9003 lbs. per sec. 

557. Transmission of Compressed Air; through very Long 
Level Pipes. Steady Flow. 

Case I. When the difference between the tensions in the 
reservoirs at the ends of the pipe is small. — Fig. 622. Under 




■■■■r 



Fig. 622. 




these circumstances it is simpler to employ the form of formula 
that would be obtained for a liquid by applying Bernoulli's 
Theorem, taking into account the " loss of head " occasioned 



TRANSMISSION OF COMPEESSED AIR. 787 

by the friction on the sides of the pipe. Since the pipe is 
very long, and the change of pressure small, the mean velocity 
in the pipe, v' f assumed to be nearly the same at all points 
along the pipe, will not be large ; hence the difference be- 
tween the velocity-heads at n and m will be neglected ; a cer- 
tain mean heaviness y' will be assigned to all the gas in the 
pipe, as if a liquid. 

Applying Bernoulli's Theorem, with friction, § 516, to the 
ends of the pipe, n and ra, we have (as for a liquid) 



2g ' y' 2g ' y' ~" d 2g 



r ^ L + Pm + = ^ L+ Pp + -4:f~~. . (1) 



Putting (as above mentioned) v m 2 — v„ = 0, we have, more 
simply, 2 

y' ~ J d'2g K) 

The value of f as coefficient of friction for air in long 
pipes is found to be somewhat smaller than for water ; see next 
paragraph. * 

558. Transmission of Compressed Air. Experiments in the St. 
Gothard Tunnel, 1878.— [See p. 96 of Vol. 24 (Feb. '81), Yan 
Xostrand's Engineering Magazine.] In these experiments, 
the temperature and pressure of the flowing gas (air) were ob- 
served at each end of a long portion of the pipe which delivered 
the compressed air to the boring-machines three miles distant 
from the tunnel's mouth. The portion considered was selected 
at a distance from the entrance of the tunnel, to eliminate the 
fluctuating influence of the weather on the temperature of the 
flowing air. A steady flow being secured by proper regulation 
of the compressors and distributing tubes, observations were 
made of the internal pressure (p), internal temperature (T), as 
well as the external, at each end of the portion of pipe con- 
sidered, and also at intermediate points ; also of the weight 
of flow per second G = Q o y , measured at the compressors 
under standard conditions (0° Cent, and one atmos. tension). 
Then "knowing the p and T at any section of the pipe, the 



788 



MECHANICS OF ENGINEERING. 



heaviness y of the air passing that section can be computed 





TA 



and the velocity v = G -5- Fy, F being 



from £- = 

L K P> 

the sectional area at that point. Hence the mean velocity v\ 
and the mean heaviness y', can be computed for this portion 
of the pipe whose diameter = d and length = I. In the ex- 
periments cited it was found that at points not too near the 
tunnel-mouth the temperature inside the pipe was always 
about 3° Cent, lower than that of the tunnel. The values of 
f in the different experiments were then computed from eq. 
(2) of the last paragraph ; i.e., 



y' 7 ^V 



(2) 



all the other quantities having been either directly observed, 
or computed from observed quantities. 

THE ST. GOTHARD EXPERIMENTS. 

[Concrete quantities reduced to English units.'] 



No. 


I 


d 


y 
Qbs. cub. 


Atmospheres. 


P n -Pm 


v' 


mean 
temp. 


/ 
















(feet.) 


(ft.) 


ft.) 


Pn 


Pm 


lbs. sq. in. 


ft. per sec. 


Cent. 




1 


15092 


t 


0.4058 


5.60 


5.24 


5.29 


19.32 


21° 


.0035 


2 


15092 


t 


0.3209 


4.35 


4.13 


3.23 


16.30 


21° 


.0038 


3 


15092 


t 


.2803 


3.84 


3.65 


2.79 


15.55 


21° 


.0041 


4 


1712 


i 


.3765 


5.24 


5.00 


3.52 


37.13 


26.5 


.0045 


5 


1712 




.3009 


4.13 


4.06 


1.03 


30.82 


26.5 


. 0024(?) 


6 


1712 


i 


.2641 


3.65 


3.54 


1.54 


29.34 


26.5 


.0045 



In the article referred to ( Yan Nostrand's Mag.) f is not 
computed. The writer contents himself with showing that 
Weisbach's values (based on experiments with small pipes and 
high velocities) are much too great for the pipes in use in the 
tunnel, 

"With small tubes an inch or less in diameter Weisbach 
found, for a velocity of about 80 ft, per second,/ =.0060; 
for still higher velocities/ was smaller, approximately, in ac- 
cordance with the relation 

f= .0542 

Vv f (in ft. per sec.) 



TRANSMISSION OF COMPEESSED AIE. 789 

On p. 370, vol. xxiv, Tan Nostrand's Mag., Prof. Kobinson 
of Ohio mentions other experiments with large long pipes. 

From the St. Gothard experiments a value oif = .004 may 
be inferred for approximate results with pipes from 3 to 8 in. 
in diameter. 

Example.— It is required to transmit, in steady flow, a supply 
of G = 6.456 lbs. of atmospheric air per second through a pipe 
30000 ft. in length (nearly six miles) from a reservoir where 
the tension is 6.0 atmos. to another where it is 5.8 atmos., the 
mean temperature in the pipe being 80° Fahr., = 24° Cent, 
(i.e. = 297° Abs. Cent.). Required the proper diameter of 
pipe ; d = ? The value f = .00425 will be used, and the f t.- 
lb.-sec. system of units. The mean volume passing per second 
in the pipe is 

Q'=G + y' (3) 

The mean velocity may thus be written : v' = -—- = -^ . (4) 

The mean heaviness of the flowing air, computed for a mean 
tension of 5.9 atmospheres, is, by § 472, 

r - iirnTr • m x - 0807 = °- 431 lbs - per eub - ft ; 

and hence, see eq. (3), 

at tension of 5.9 atmos., and temperature 297° Abs. Cent. 
Now, from eq. (2), 



whence 



v = Al. y' 1 £° . (5) 



790 



MECHANICS OF ENGINEEKING. 



and hence, numerically, 



d = 



5 / 4 X .00425 X 0.431 X 30000 X (14. 74) 2 _ 
Y (.7854) 2 [14.7 X 144(6.00 - 5.80)]2 X 32.2 " 



,23 feet 



559. (Case II of § 557) Long Pipe, with Considerable Differ- 
ence of Pressure at Extremities of the Pipe. Flow Steady. — Fig. 
623. If the difference between the end-tensions is compara- 
tively great, we can no longer deal with the whole of the air 





*vJ.\ • m. 



Fig. 623. 



in the pipe at once, as regards ascribing to it a mean velocity 
and mean tension, but must consider the separate laming 
such as AB (a short length of the air-stream) to which we may 
apply eq. (2) of § 557 ; A and B corresponding to the n and 
m of Fig. 622. Since thep n —p m , I, y' , and v' of § 557 
correspond to the — dp, ds, y, and v of the present case (short 
section or lamina), we may write 



V J dig 



(1) 



But if G = weight of flow per unit of time, we have at any 
section, Fvy = G (equation of continuity) ; i.e., v = G -i- Fy, 
whence by substitution in eq. (1) we have 



dp___±f_ G*ds , 
~y~~2gd 'Fy 2 ' 



i.e., — ydp = 



4:fG 2 

2gF 2 d 



ds. 



(2) 



Eq. (2) contains three variables, y, p, and s (= distance of 
lamina from n f ). As to the dependence of the heaviness y on 
the tension^? in different laminae, experiment shows that in most 
cases a uniform temperature is found to exist all along the 
pipe, if properly buried, or shaded from the sun ; the loss of 
heat by adiabatic expansion being in great part made up by 
the heat generated by the friction against the walls of the 



GAS IN LONG PIPES. LAEGE FALL OF TENSION. 791 

pipe. This is due to the small loss of tension per unit of 
length of pipe as compared with that occurring in a short dis- 
charge pipe or nozzle. Hence we may treat the now as iso- 
thermal, and write j? ~ y = p n >-i- y n > (§ 475, Mariotte's Law). 

Hence y = — p, which substituted in eq. (2) enables us to 

•••-XS*=G^]i"* < 8 » 

Performing the integration, noting that at n' p =p n f > s = 0, 
and atm'j?= p m , and s = I, we have 

jLj- » _ a-i ±f}_ (?_ Prs j isothermal flow ) ,.. 

2iPn' Pm J g ^ ' F 2 ' y n >' ' { in long pipes ) • " w 

It is here assumed that the tension at the entrance of the pipe 
is practically equal to that in the head reservoir, and that at 
the end (m f ) to that of the receiving reservoir; which is not 
strictly true, especially when the corners are not rounded. It 
will be remembered also that in establishing eq. (2) of § 557 
(the basis of the present paragraph), the "inertia" of the gas 
was neglected ; i.e., the change of velocity in passing along 
the pipe. Hence eq. (4) should not be applied to cases where 
the pipe is so short, or the difference of end-tensions so great, 
as to create a considerable difference between the velocities at 
the two ends of the pipe. (See Addendum on p. 797.) 

Example. — A well or reservoir supplies natural gas at a ten- 
sion of p n > = 30 lbs. per sq. inch. Its heaviness at 0° Cent, 
and one atmosphere tension is .0484 lbs. per cub. foot. In 
piping this gas along a level to a town two miles distant, a 
single four-inch pipe is to be employed, and the tension in the 
receiving reservoir (by proper regulation of the gas distributed 
from it) is to be kept equal to 16 lbs. per sq. in. (which would 
sustain a column of water about 2 ft. in height in an open 
water manometer, Fig. 465). 



792 



MECHANICS OF ENGINEEEING. 



The mean temperature in the pipe being 17° Cent., required 
the amount (weight) of gas delivered per second, supposing 
leakage to be prevented (formerly a difficult matter in practice). 
Solve (4) for G, and we have 



G 



= wv/ 



gd Yn 



(Pn* -i?m' 2 )' 



(5) 



First, from § 472, with T n , = T m , = 290° Abs. Cent., we 
compute 



Yn' y T« 
Hence with/* = .005, 



14.7 X 144 290 
.0485 - ' 273 



46454 feet. 



2 X A[(30 X 1 44 ) 2 -(16 X 144) a ] 
4X.005X 10560x46454 



= 0.337 lbs. per sec. 

(For compressed atmospheric air, under like conditions, we 
would have G = 0.430 lbs. per second.) 

Of course the proper choice of the coefficient f has an im- 
portant influence on the result. 

From the above result (G = 0.337 lbs. per second) we can 
compute the volume occupied by this quantity of gas in the 

receiving reservoir, using the relation Q m > = — . 

Ym' 

The heaviness y m > of the gas in the receiving reservoir is 
most easily found from the relation -t^- = ±-^ , which holds 

Ym' Yn' 

good since the flow is isothermal. I.e., — = 46454 ft.; 

Ym' 

whence y m > — 0.049 lbs. per cubic foot, p m > being 16 X 144 
lbs. per sq. ft. 
Hence 

^' = -£: = SS = 6 - Y94cub - ft -P ersec - 



FLOW OF GAS IN PIPES. 793 

It should be said that the pressure at the up-stream end of 
the pipe depends upon the rate of flow allowed to take place. 

With no flow permitted, the pressure in the tube of a gas- 
well has in some cases reached the high figure of 500 or 600 
lbs. per sq. in. 

560. Rate of Decrease of Pressure along a Long Pipe. — Con- 
sidering further the case of the last paragraph, that of a 
straight, long, level pipe of uniform diameter, delivering gas 
from a storage reservoir into a receiving reservoir, we note 
that if in eq. (4) we retain p m > to indicate the tension in the 
receiving reservoir, but let p n > denote in turn the tension at 
points in the pipe successively further and further (a distance 
x) from the receiving reservoir m', we may write x for I and 
obtain the equation (between two variables, jv and x) 

Pn' 2 —Pm? -■ Const. Xx (6) 

This can be used to bring out an interesting relation men- 
tioned by a writer in the Engineering News of July 1887 
(p. 71), viz., the fact that in the parts of the pipe more distant 
from the receiving end, m', the distance along the pipe in 
which a given loss of pressure occurs is much greater than 
near the receiving end. 

To make a numerical illustration, let us suppose that the 
. pipe is of such size, in connection with other circumstances, 
that the tension p n > at A, a distance x == six miles from m', is 
two atmospheres, the tension in the receiving reservoir being 
one atmosphere ; that is, that the loss of tension between A 
and m' is one atmosphere. If we express tensions in atmos- 
pheres and distances in miles, we have for the value of the 
constant in eq. (6), for this case, 

Const. = (4 — 1) -^ 6 = | ; {for assumed units.) . . (7) 

Now let jv = the tension at B, a point 18 miles from m\ 
and we have, from eqs. (6) and (7), the tension at B = 3.16 
atmospheres. Proceeding in this manner, the following set of 
values is obtained : 



794 



MECHANICS OF ENGINEERING. 



Point. 


Total distance 
from mf. 


Distance be- 
tween consecu- 
tive points. 


Tension at 
point. 


Loss of ten- 
sion in each in- 
terval. 


F 


126 miles. 


36 miles. 


8.00 aim. 


1.22 aim. 


E 


90 " 


30 " 


6.78 " 


1.22 " 


D 


60 " 


24 " 


5.56 " 


1.21 " 


C 


36 " 


18 " 


4.35 " 


1.19 " 


B 


18 " 


12 " 


3.16 " 


1.16 " 


A 


6 " 


6 " 


2.00 " 


1.00 " 


m! 


" 




1.00 " 









If the distances and tensions in the second and fourth 
columns be plotted as abscissae and ordinates of a curve, the 
latter is a parabola with its axis following the axis of the pipe ; 
its vertex is not at m\ however. 

561. Long Pipe of Variable Diameter. — Another way of stat- 
ing the fact mentioned in the last paragraph is as follows : At 
the up-stream end of the pipe of uniform diameter the gas is 
of much greater density than at the other extremity (the 
heaviness is directly as the tension, the temperature being as- 
sumed the same throughout the pipe), and the velocity of its 
motion is smaller than at the discharging end (in the same 
ratio). It is true that the frictional resistance per unit of 
length of pipe varies directly as the heaviness [eq. (1), § 510], 
but also true that it varies as the square of the velocity ; so 
that, for instance, if the pressure at a point A is double that 
at B in the pipe of constant diameter, it implies that the 
heaviness and velocity at A are double and half, respectively, 
those at B, and thus the gas at A is subjected to only half the 
frictional resisting force per foot of length as compared with 
that at B. Hence the relatively small diminution, per unit of 
length, in the tension at the up-stream end in the example of 
the last paragraph. 

In the pipe of uniform diameter, as we have seen, the greater 
part of the length is subjected to a comparatively high ten- 
sion, and is thus under a greater liability to loss by leakage 
than if the decrease of tension were more uniform. The 
total u hoop-tension^ (§ 426) in a unit length of pipe, also, is 
proportional to the gas tension, and thinner walls might be 
employed for the down-stream portions of the pipe if the gas 



FLOW OF GAS IN PIPES. 795 

tension in those portions could be made smaller than as shown 
in the preceding example. 

To secure a more rapid fall of pressure at the up-stream end 
of the pipe, and at the same time provide for the same delivery 
of gas as with a pipe of uniform diameter throughout, a pipe 
of variable diameter may be employed, that diameter being 
considerably smaller at the inlet than that of the uniform pipe 
but progressively enlarging down-stream. This will require 
the diameters of portions near the discharging end to be larger 
than in the uniform pipe, and if the same thickness of metal 
were necessary throughout, there would be no saving of metal, 
but rather the reverse, as will be seen ; but the diminished 
thickness made practicable in those parts from a less total hoop 
tension than in the corresponding parts of the uniform pipe 
more than compensates for the extra metal due to increased 
circumference, aside from the diminished liability to leakage, 
which is of equal importance. 

A simple numerical example will illustrate the foregoing. 
The pipe being circular, we may replace F by \nd 2 in equation 
(4), and finally derive, G being given, 



UV — i>m' J Liv — JV J 



(8). 



Let A be the head reservoir, and m / the receiving reservoir, 
and B a point half-way between. At A the tension is 10 at- 
mospheres ; at m', 2 atmospheres. For transmitting a given 
weight of gas per unit-time, through a pipe of constant diam- 
ter throughout, that diameter must be (tensions in atmospheres ; 
2l being the length), by eq. (8), 



d= CI. 



07 O *(.O2O8)» = 0.46 CI*. . (8) x 



If we substitute for the pipe mentioned, another having a con- 
stant diameter d 1 from A to B, where we wish the tension to 
be 5 atmospheres, and a different constant diameter d % from B 
to m', we derive similarly 



796 



MECHANICS OF ENGINEEKING. 



and 



d, = CI* 



j5^i] W54OT 



It is now to be noted that the sum of d x and d 2 is slightly 
greater than the double of d ; so that if the same thickness of 
metal were used in both designs the compound pipe would 
require a little more material than the uniform pipe; but, 
from the reasoning given at the beginning of this paragraph, 
that thickness may be made considerably less in the down- 
stream part of the compound pipe, and thus economy secured. 

[In case of a cessation of the flow, the gas tension in the 
whole pipe might rise to an equality with that of the head- 
reservoir were it not for the insertion, at intervals, of auto- 
matic regulators, each of which prevents the decrease of ten- 
sion on its down-stream side below a fixed value. To provide 
for changes of length due to rise and fall of temperature, the 
pipe is laid with slight undulations.] 

It is a noteworthy theoretical deduction that a given pipe of 
variable diameter connecting two reservoirs of gas at specified 
pressures will deliver the same weight of gas as before, if 
turned end for end. This follows from equation (3)', § 559. 
With d variable, (3)' becomes (with F— \nd*) 



ds 
oT> 



i.e., G 2 = - Pn ' $% - 



(9) 






(C" is a constant.) 

r m 'ds 

But / , — is evidently the same in value if the pipe be 

turned end for end. In commenting on this circumstance, we 
should remember (see § 559) that the loss of pressure along the 
pipe is ascribed entirely to frietional resistance, and in no de- 
gree to changes of velocity (inertia). 

On p. 73 of the Engineering News of July 1887 are given 
the following dimensions of a compound pipe in actual use, 
and delivering natural gas. The pressure in the head-reservoir 
is 319 lbs. per sq. in.; that in the receiving reservoir, 65. For 
-2.84 miles from the head-reservoir the diameter of the pipe is 



NATURAL GAS. COEFFICIENT OF FLUID FRICTION. 797 

8 in. ; throughout the next 2.75 miles, 10 in. ; while in the 
remaining 3.84 miles the diameter is 12 in. At the two 
points of junction the pressures are stated to be 185 and 132 
lbs. per sq. in., respectively, during the flow of gas under the 
conditions mentioned. 



561a. Values of the Coefficient of Fluid Friction for Natural 
Gas. — In the Ohio Report on Economic Geology for 1888 may 
be found an article by Prof. S. W. Robinson of the University 
of that State describing a series of interesting experiments 
made by him on the flow of natural gas from orifices and 
through pipes. By the insertion of Pitot tubes approximate 
measurements were made of the velocity of the stream of gas 
in a pipe. The following are some of the results of these ex- 
periments, jp x —p 2 representing the loss of pressure (in lbs. per 
sq. inch) per mile of pipe-length, and f the coefficient of fluid 
friction, in experiments with a six-inch pipe : 



Vx-V-x 


1.00 


1.50 


2.25 


2.50 


5.75 


6.25 


f 


0.0025 


0.0037 


0.0052 


0.0059 


0.0070 


0.0060 



In the flow under observation Prof. Robinson concluded 
that f could be taken as approximately proportional to the 
fourth root of the cube of the velocity of flow ; though calling 
attention to the fact that very reliable results could hardly be 
expected under the circumstances. 

561b. [Addendum to § 559.] Isothermal Flow in a level pipe, 
with consideration of Inertia. — In eq. (1) of p. 697 neglect dz, 
put w -T- F — 4 -r- d, and divide through by v\ In the second 
term put G 2 -t- F*y> for v* and then^> (y m +p m ) for y. We 
now find the variables separated, and on integration for steady 
flow obtain (after putting v n -=- v m = Fy m -f- Fy n =p m -j-i?n)> 

^Pmi 2 ' G 2 ' p m d' 

[Notation as in § 557 with G = Fvy.] 



MS - 



CHAPTEE IX. 



IMPULSE AND RESISTANCE OF FLUIDS. 



562. The so-called "Reaction" of a Jet of Water flowing from 
a Vessel. — Id Fig. 624, if a frictionless but water-tight plug B 

be inserted in an orifice in the 
vertical side of a vessel mounted 
on wheels, the resultant action of 
the water on the rigid vessel (as a 
whole) consists of its weight G, 
and a force P' — Fhy (in which 
F= the area of orifice) which is 
the excess of the horizontal hydro- 
static pressures on the vessel wall 
toward the right ( || to paper) over 
those toward the left, since the 
pressure P, = Fhy, exerted on the plug is felt by the post C, 
and not by the vessel. Hence the post D receives a pressure 




Fig. 624. 



P' - Fhy. 



(1) 



Let the plug B be removed. A steady flow is then set up 
through the orifice, and now the pressure against the post D is 
2Fhy (as will be proved in the next paragraph) ; for not only 
is the pressure Fhy lacking on the left, because of the orifice, 
but the sum of all the horizontal components ( || to paper) of 
the pressures of the liquid filaments against the vessel wall 
around the orifice is less than its valu^ before the flow began, 
by an amount Fhy. A resistance B = 2Fhy being provided, 
and the post removed, a slow uniform motion may be main- 
tained toward the right, the working force being 2Fhy == P" 

798 



"be action" of a jet. 



799 



y^= 




Fig. 625. 



(see Fig. 625 ; R is not shown). If an insufficient resistance 

be furnished before removing the post Z>, . . . . „„ ., 

the vessel will begin to move toward the 

right with an acceleration, which will 

disturb the surface of the water and 

change the value of the horizontal force. 

This force 

P // = 2Fhy ... (2) 

is called the " reaction'' of the water-jet ; 
y is the heaviness of the liquid (§ 7). 

Of course, as the flow goes on, the 
water level sinks and the t% reaction" diminishes accordingly. 
Looked upon as a motor, the vessel may be considered to be a 
piston-less and valve-less water-pressure engine, carrying its 
own reservoir with it. 

In Case II of § 500 we have already had a treatment of the 
" Reaction- wheel " or "Barker's mill," which is a practical 
machine operating on this principle, and will be again con- 
sidered in " Notes on Hydraulic Motors." 

563. " Reaction" of a Liquid Jet on the Vessel from which it 
Issues. — Instead of showing that the pressures on the vessel 
close to the orifice are less than they were when there was no 
flow by an amount Fhy (a rather lengthy demonstration), 
another method will be given, of greater simplicity but some- 
what fanciful. 

If a man standing on the rear platform of a car is to take up 
in succession, from a basket on the car, a number of balls of 
equal mass = M, and project each one in turn horizontally 
backward with an acceleration =J>, he can accomplish this 
only by exerting against each ball a pressure = Mp, and in the 
opposite direction against the car an equal pressure = Mp. If 
this action is kept up continuously the car is subjected to a 
constant and continuous forward force of P" = Mp. 

Similarly, the backward projection of the jet of water in the 
case of the vessel at rest must occasion a forward force against 
the vessel of a value dependent on the fact that in each small 
interval of time At a small mass AM of liquid has its velocity 
changed from zero to a backward velocity of v = V%gh ; that 






800 MECHANICS OF ENGINEERING. 

is, has been projected with a mean acceleration of p = , 

At 
so that the forward force against the vessel is 

P" — mass X ace. = — (3). 

At 

If Q = the volume of water discharged per unit time, then 
AM — -^ At, and since also Q = Fv = FV2gh, eq. (3) be- 

comes "Reaction" of jet =P' f = ZFhy. . . . (4) 

(A similar proof, resulting in the same value for P" , is 
easily made if the vessel has a uniform motion with water sur- 
face horizontal.) 

If the orifice is in " thin plate," we understand by F the 
area of the contracted section. Practically, we have v= V2gh 
(§ 495), and hence (3) reduces to 

P" = 2<P*Fhy (5) 

Weisbach mentions the experiments of Mr. Peter Ewart of 
Manchester, England, as giving the result P" = 1.73Fhy 
with a well-rounded orifice as in Fig. 625. He also found 
(p = .94 for the same orifice, so that by eq. (4) we should have 

P" = 2(MyFh r = 1.77Fhy. 

With an orifice in thin plate Mr. Ewart found P" = 
1.14:Fhy. As for a result from eq. (4), we must put, for F, 
the area of the contracted section SiF (§ 495), which, with 
= .96, gives 

P" = 2(.my.64:Fhy = 1.18Fhy. ... (6) 

Evidently both results agree well with experiment. 

Experiments made by Prof. J. B. Webb at the Stevens 
Institute (see Journal of the Franklin Inst., Jan. '88, p. 35) 
also confirm the foregoing results. In these experiments the 
vessel was suspended on springs and the jet directly down- 
ward, so that the "reaction" consisted of a diminution of the 
tension of the springs during the flow. 

564. Impulse of a Jet of Water on a Fixed Curved Vane (with 
Borders). — The jet passes tangentially upon the vane. Fig. 



IMPULSE OF JET. 



801 






tt^Jk 




626. B is the stationary nozzle from which a jet of water of 

cross-section i^(area) and velocity = c impinges tangentially 

upon the vane, which has 

plane borders, parallel to 

paper, to prevent the lat- 
eral escape of the jet. 

The curve of the vane is 

not circular necessarily. 

The vane being smooth, 

the velocity of the water 

in its curved path remains 

= c at ail points a*ong 

the curve. Conceive the 
curve divided into a great 
number of small lengths each = ds, and subtending some 
angle = dcp from its own centre of curvature, its radius of 
curvature being = r (different for different ds'a), which makes 
some angle = with the axis Y ( "J to original straight jet 
BA). At any instant of time there is an arc of water AD m 
contact witli the vane, exerting pressure upon it. The pres- 
sure dP of any ds of the vane against the small mass of water 
Fds .y-±g then in contact with it is the " deviating" or " cen- 
tripetal " force accountable for its motion in a curve of radius 
= r, and hence must have a value 



Fig. 626. 



dP 



The opposite and equal of this force is the dP shown in 
Fig. 614, and is the impulse or pressure of this small mass 
against the vane. Its Xcomponent is dX = dP sin 0. By 
making vary from to a, and adding up the corresponding 
values of dX j we obtain the sum of the ^-components of the 
small pressures exerted simultaneously against the vane by the 
arc of water then in contact with it ; i.e., noting that <fe=rd0 y 



TdX^tdP . sin 4, = ill /- <fe-ri"0 



g 

Fyc 2 




1 f a Pvc 2 r a 

■Jo [ sin 0]^0 = - L -\ - cos 0; 
9 Lo 



802 



MECHANICS OF ENGINEEBING. 



hence the X-tmpulse ) Fye n 1 Qyc ri ., /ON 

, n 7 ± \ = — — 1— cos a] = -^— 1 — cos a], (2) 

\ fixed vane ) n L <7 J> w 



against y 



in which Q = Fc = volume of water which passes through the 
nozzle (and also = that passing over the vane, in this case) per 
unit of time, and a == angle between the direction of the 
stream leaving the vane (i.e., at D) and its original direction 
(BA of the jet) ; i.e., a = total angle of deviation. Similarly, 
the sum of the i^-components of the dP's of Fig. 626 may be 
shown to be 

Y-invpulse on fixed vane = / dP . cos = -^L sin or...(2)' 
Hence the resultant impulse on the vane is a force 



QyG 



P" = VX* + Y* = ^ 4/2(1- cos a), 

id 



(3) 



and makes such an angle a', Fig. 627, with the direction BA> 
that 



. Y mi a 

tan a' = -= = . 

X 1 — cos a 



(4) 



©M 





Fig. 627. 



Fig. 628. 



For example, if * = 90°, then <*' =s 45°; while if a = 180°, 
Fig. 628, we have a! = 0° ; i.e., P" is parallel to the jet iLi, 
and its value is 



P"=2Qy- 



IMPULSE OF JET ON VANE. 80 g 

curved vane, with borders, of the pre! 

ceding paragraph be replaced by a e 

80 hd of revelation, Fig. 629, with its 0&- 

Z; n "? ,° f tLe ** ^'resultan 
pressure of the jet upon it will simply 

be the sum of the ^-components CiJ 
- to£A) of the pressures on all ele 
ments of the surface at => m v ■ \. F,G - 6S9 

uxldce at a given instant; i.e. 







«0 



pulse" (with or = 90°), 




lSS^'n e T rime , ntS ° f Kdone ' ™de in 
1838 confirm the truth of eq. (6) quite 

I**.- ' J* T 7 ' ^ d ° alS ° th ° Se 0f ^o studeul of 
delphia (see /« -VSK^iS?^ S™* 

M- (6) is apphcable to the theo- ' P- '' 

rj of Pitot's Tube (see § 539) 
*ig. 631, if we consider the edge 

of the tube plane and quite wide. 
The water in the tube is at rest, 
and its section at A (of area= M 

may be treated as a flat vertical 
plate receiving not only the 
hydrostatic pressure Iky, due 
to the depth x below the sur- 
face, but a continuous impulse P" - Ft > T 

p ^ -Hey + g [see eq. (6)J 




MECHANICS OF ENGINEEBIHG. 



. f the end A, of the stationary column 

Forthe equilibrium of the end A, 
AD, we must have, therefore, 

*• „ fn corresponds reasonably well 
The relation * «^T $ experiments with the instru- 
ct the results of Vj^X^f, on trial of an in- 
ment mentioned in « BM- J ^ at ^ we re made flar- 



h' = (1-5) 



(7)' 



2? 



, , . f t]ie tube should occasion 
while Darcy, desirous that the e«a surrounding stream 

* -e *— e ^^^ —lly 

convergent* The law 

relation , 

y = almost exactly (1.0) gr- ■ CO 

(See §539.) tf is made cup- 

If the sol id ot revo ve (ag m 

gna ped, as in Fig; 632 - ^ 

Fig. 628) « = 1W > ana 

eq. (5), 




Fig. 632. 






(8) 



so ft per sec. and the jet 
Ex.^.-Bg.;32. Kj=f° ;^ e Uquid bei ng water, 

(cylindrical) has ******* £ we have [ft, lb., sec] 
so that r = 62.5 lbs. per cub. ft., 

„5fiV900X62.5 
^U2i______- -= 19.05 lbs. 

the impulse (force) = P" = ' 32.2 

Wnrobablyshowasmallerresult. 

Experimentwmnd^robaoy^^^^^-- 



IMPULSE OF JET ON MOVING VANE. 



805 




Fig. 633. 



566. Impulse of a liquid Jet upon a Moving Vane having 
lateral Borders and Moving in the Direction of the Jet.— Fig. 
633. The vane has a motion of translation (§ 108) in the 
same direction as the jet. Call this the axis X. It is moving 
with a velocity v away from the jet (or, if toward the jet, v 
is negative). We con- 
sider v constant, its ac- 
celeration being prevented 
by a proper resistance 
(such as a weight = G) 
to balance the JT-com- 
ponents of the arc-pres- 
sures. Before coming in 
contact with the vane, 
which it does tangentially 
(to avoid sudden devia- 
tion), the absolute velocity 
(§ 83) of the water in the 
jet = c, while its velocity 
relatively to the vane at A is = o - v.; and it will now be 
proved that the relative velocity along the vane is constant. 
bee Eig.^ 634. Let v = the velocity of the vane (of each 
point of it, since its motion is one of translation), and u = the 
velocity of a water particle (or small mass of water of length 
= ds) relatively to the point of the vane which it is passing 
Then w, the absolute velocity of the small mass, is the diago- 
nal formed on u and t>. Neglecting friction, the only actual 
torce acting on the mass is P, the pressure of the vane against 
it and this is normal to the curve. Now an imaginary system 
of forces, equivalent to this actual system of one force P, i.e., 
capable of producing the same motion in the mass, may be 
conceived of, consisting of the individual forces which would 
produce, separately, the separate motions of which the actual 
motion of this small mass M is compounded. These com- 
ponent motions are as follows : 

1. A horizontal uniform motion of constant velocity^- 

and J ' 

2. A motion in the arc of a circle of radius = r and with a 



806 MECHANICS OP ENGINEEKING. 

velocity = u, which we shall consider variable until proved 
otherwise. 

Motion 1 is of such a nature as to call for no force (by New- 
ton's first law of motion), while motion 2 could be maintained 

Mu 2 
by a system of two forces, one normal, P w , = , and the 

du 
other tangential, P t = M — [see eq. (5), p. 76]. This imagi- 
ne 

nary system of forces is shown at (II.), Fig. 634, and is equiv- 



(ii.) i 
\0 




u v r -, 

*~- -[_ U ° = w ~' v J 

Fig. 634. 

alent to the actual system at (I.). Therefore ,2 (tang, com- 
pons.) in (I t ) should be equal to 2 (tang, compons.) in (II.) ; 
whence we have 

P, = 0; i.e., ** = 0; or J = 0; . . (!) 

i.e., u is constant along the vane and is equal to c — v at every 
point. (The weight of the mass has been neglected since the 
height of the vane is small.) In Fig. 634 the symbol w has 
been used instead of c, and the point corresponds to A in 
Fig. 633. 

[N.B. If the motion of the vane were rotary, about an axis 
•~] to AB (or to c), this relative velocity would be different at 
different points. See Notes on Hydraulic Motors. If the 
radius of motion of the point A, however, is quite large com- 
pared with the projection of AD upon this radius, the relative 
velocity is approximately = c — v at all parts of the vane, 
and will be taken =c— v in treating the " Hurdy gurdy" in 
§ 567.] 



WORK OF JET ON VANE. 807 

By putting 2 (normal compons.) of (I.) == 2 (normal com- 
pons.) in (II.) we have 

P = P n ; i.e., P = M^-= J ^°-^ ; ... (2) 

r r 

so that to find the sum of the X-components of the pressures 
exerted against the vane simultaneously by all the small masses 
of water in contact with it at any instant, the analysis differs 
from that in § 564 only in replacing the c of that article by 
the (c — v) of this. Therefore 

Fy 

2(X-pressures) = P x = — - (c — v)*[l — cos a], . (3) 

(where a is the angle of total deviation, relatively to vane, of 
the stream leaving the vane, from its original direction), and 
is seen to be proportional to the square of the relative velocity. 
i^is the sectional area of jet, and y the heaviness (§ 7) of the 
liquid. The ^-component (or P Y ) of the resultant impulse 
is counteracted by the support FF, Fig. 633. Hence, for a 
uniform motion to he maintained, with a given velocity = v, 
the weight G must be made = P x of eq. (3). (We here 
neglect friction and suppose the jet to preserve a practically 
horizontal direction for an indefinite distance before meeting 
the vane. If this uniform motion is to be toward the jet, v 
will be negative in eq. (3), making P x (and .'. G) larger than 
for a positive v of same numerical value. 

As to the doing of work [§§ 128, etc.], or exchange of 
energy, between the two bodies, jet and vane, during a uni- 
form motion away from the jet, P x exerts a power of 

Z = P x v = ?L (o - vfv\l - cos a], . . . (4) 

if 

in which L denotes the number of units of work done per unit 
of time by P x \ i.e., the power (§ 130) exerted by P x . 
If v is negative, call it — v\ and we have the 



MECHANICS OF ENGINEEKING. 



808 

,. ii r WP ar e more concerned with eq. (4) 

Of course, practically, we are ujuio v, 

Ot course v j fa & maximum for „ = £ c 

than with (5). l&e powei w eannot 

work is being done on the hinder one , i.e., 
utilized as fast as it issues from the nozzle. 

vane, in eq. (4) ; whence 



Power exerted on 
series of vanes 



-r=«iti- 



,- 



jj-* ^ul. [1 — cos a](c — v)v. 
9 



(6) 



i *4.:««. ^7 T'-^rlw—O. whence c— 2^=0, 

Kg .«,w,h , P-^"* 1 ,^ „ „„e, ,. .be 
maximum power, 

Q'y <? M'e\ ( «=180°, 1 ..(7) 

in which Jf ' denotes the mass of the flow 
per unit of time from the stationary 

nozzle. Now *£ is the «*« i«* 

enerav furnished per unit of time by the 
et hence the niotor of Fig. 635 (~ 

5i has a theoretical efficiency of 
fig. 635. of cups) nas a . g 

nnity, utilizing all the Irinefcc £-»££ ^ where the y 

true, the absolute veloci y o the parti ^ ^ ^ ^ 

leave the cup, or vane, should be M»t w 




POWER OF JETS. qqq 

tSvZ S L MJT '- 0rJ1 '' the Vel0dty0f the P articl - rela- 
ttyely to the vane la = o-v= what it was at A, and hence 

1S - ° ~2 = h hence at E the absolute velocity is «,= 
(rel. veloc. £ toward left)-(veloc. £ of vane toward right) = ; 
■Q.E.D. For v > or < ^ this efficiency will not he attained. ' 

567. The California "Hurdy-gurdy;" or Pelton Wheel -The 
■efficiency of unity in the series of cups just mentioned is in 
practice reduced to 80 or 85 per cent from friction Td to" 
escape of water. The 
Pelton wheel or Cali- 
fornia " Hurdy-gur- 
dy," shown (in prin- 
ciple only) in Fig. 636, \ 
is designed to utilize \ 
the mechanical rela- 
tion just presented, 

and yields results con- f 7pJ 

firming the above the- *- ^ 

ory, viz., that with the 
Jinear velocity of the 




Fig. 636 



cnp-centres regulated to equal £, and with a = 180°, the effi. 
ciency approaches unity or 100 per cent. Each cup has a pro 
^ectmg sharp edge or rib along the middle, to split thejlt 7^ 

TO i?S Wh A el - WaS iDVented t0 Utilize sma]1 Jets of very great 
gu o is great, still, by giving a large value 
to r, the radius of the wheel, the making of „ = £ does not 
necessitate an inconveniently great speed of roiion (i e 
revelations per unit of time). The plane of the wheel mav' 
be in any convenient position. J 

In the London Engineer of Mav '84 n W7 ,•„ • 



810 



MECHANICS OF ENGINEERING. 



showed an efficiency of 87 per cent. The diameter of the 
wheel was only 6 ft., that of the jet 1.89 id., and the head of 
the supply reservoir 386 ft., the water being transmitted 
through a pipe of 22 inches diameter and 6900 ft. in length. 
107 H. P. was developed by the wheel. 

Example. — If the jet in Fig. 636 has a velocity c = 60 ft. 
per second, and is delivered through a 2-inch nozzle, the total 
power due to the kinetic energy of the water is (ft., lb., sec.) 

^ -^= 3 4,-f(a 2 X60x6 2 .5xiX3600=4566.9 ) *J£ 

and if, by making the velocity of the cups = - = 30 ft. per 

sec, 85 per cent of this power can be utilized, the "oovver of 
the wheel at this most advantageous velocity is 

L = .85 X 4566.9 = 3881 ft. lbs. per sec. = 7.05 horse-power 

[since 3881 -r- 550 = 7.05] (§ 132). For a cup-velocity of 30 
ft. per sec, if we make the radius, r, = 10 feet, the angular 
velocity of the wheel will be go = v -=- r = 3.0 radians per 
sec. (for radian see Example in § 428 ; for angular velocity, 
§ 110), which nearly = 7r, thus implying nearly a half -revolu- 
tion per sec 

568. Oblique Impact of a Jet on a Moving Plate having 

no Border. — The plate 
has a motion of trans- 
lation with a uniform 
veloc = v in a direc- 
tion parallel to jet, 
whose velocity is = c. 
At O the filaments of 
liquid are deviated, so 
that in leaving the plate 
their particles are all 
V found in the moving 
fig. 637. plane BB' of the plate 

surface, but the respective absolute velocities of these particles 







OBLIQUE JET AND PLATE. 811 

depend on the location of the point of the plate where they 
leave it, being found by forming a diagonal on the relative 
veloc. c — v and the velocity v of the plate. For example, at 
B the absolute velocity of a liquid particle is 



w = BE = Vv 2 + (c — v) 2 + 2v(c — v) cos a, 
while at B' it is 



w' = B'E' = Vv 2 + (e- v) 2 - 2v(c - v) cos a ; 

but evidently the component 1 to plate (the other component 
being parallel) of the absolute velocities of all particles leaving 
the plate, is the same and = v sin a. The skin-friction of the 
liquid on the plate being neglected, the resultant impulse of 
the jet against the plate must be normal to its surface, and its 
amount, P, is most readily found as follows : 

Denoting by AM the mass of the liquid passing over the 
plate in a short time At, resolve the absolute velocities of all 
the liquid particles, before and after deviation, into com- 
ponents 1 to the plate (call this direction Y) and || to the 
plate. Before meeting the plate the particles composing AM 
have a velocity in the direction of Y of c y = c sin a ; on leav- 
ing the plate a velocity in direction of T of v sin a : they have 
therefore lost an amount of velocity in direction of Y = 
(e — v) sin a in time At ; i.e., they have suffered an average 
retardation (or negative acceleration) in a Indirection of 

( neg. accelera- ) (c — v) sin a M x 

A= J tion H to Y\ = -J t •• • • CD 

Hence the resistance in direction of Y(i.e., the equal and op- 
posite of P in figure) must be 

A H/f 

P T = mass X 1^-accel. = — — (c— v) sin a ; . . (2) 
and therefore, since — — = M= -£2l = mass of liquid passing 

if 



812 MECHANICS OF ENGINEEKING. 

over the plate per unit of time (not that issuing from nozzle), 
we have 

^i;r1 =P=f(.-t»-n«=f (c _,)> s i™,(3) 

in which F — sectional area of jet before meeting plate. 

[N.B. Since eq. (3) can also be written P = Me sin a — 
Mo sin a, and Me sin a may be called the I^-momentum before 
contact, while Mv sin a is the T'-momentum after contact (of 
the mass passing over plate per unit of time), this method may 
be said to be founded on the principle of momentum which is 
nothing more than the relation that the accelerating force in 
any direction = mass X acceleration in that direction ; e.g., 
P x = Mp x ; P y = Mp y ; see § 74.] 

If we resolve P, Fig. 637, into two components, one, P', \\ 
to the direction of motion (v and c), and the other, P'\ ~| to 
the same, we have 

P'^Psino^-^^-^sin 2 *, . , . . (4) 



and 



Oy 
P"= P cos a = -£L- (p -— v) sin a cos a. . . (5) 



(Q = F(c — v)= volume passing over the plate per unit of 
time.) The force P" does no work, while the former, P', 
does an amount of work P'v per unit of time ; i.e., exerts a 
jpower (one plate) 

= Z = P , v=®Y(c-v)vsm*a. . . . (6) 
J/ 

If, instead of a single plate, a series of plates^ forming a 
regular succession, is employed, then, as in a previous paragraph, 
we may replace Q, = F(c — v), by Q f = Fc, obtaining as the 

Power exerted by jet ) T , Fey , . . a /m 

on series of plltes \ = L = ~f< ~ " )v Sln «• ■ W 



IMPULSE OF JET ON OBLIQUE PLATE. 818 



For v = - and a = 90° we have 



T , _lFcyc* _lM'c' 

-272-2X (8) 



max. 



= only half the kinetic energy (per time-unit) of the jet. 

569. Rigid Plates Moving in a Fluid, Totally Submerged. 
Fluid Moving against a Fixed Plate. Impulse and Resistance. — 
If a thin flat rigid plate have a motion of uniform translation 

with velocity == v through a fluid — .. 

which completely surrounds it, Fig. ~- — ~~ ^^^^5\^ 
638, a resistance is encountered (which ZZZ> \f(^{f^'inj§w\ v^ 
must be overcome by an equal and op- — ^ y^i^x^\>^- 
posite force, not shown in figure, to /7r^ '£. $&—-£-> ~^ 
preserve the uniform motion) consist- A f} ^ .yX/^^^zz^^z^ 
ing of a normal component JV, "1 to ^x^^ " 

plate, and a (small) tangential com- ^^=^^ : ^ ~~~l 

ponent, or skin-friction, T, \\ to plate. fig. 638. 

Unless the angle a, between the surface of plate and the direc- 
tion of motion ... v, is very small, i.e. unless the plate is 
moving nearly edgewise through the fluid, N is usually much 
greater than T. The skin-resistance between a solid and a fluid 
has already been spoken of in § 510. 

When the plate and fluid are at rest the pressures on both 
sides are normal and balance each other, being ordinary static 
fluid pressures. When motion is in progress, however, the 
normal pressures on the front surface are increased by the 
components, normal to plate, of the centrifugal forces of the 
curved filaments (such as AB) or "stream-lines," while on 
the back surface, D, the fluid does not close in fast enough to 
produce a pressure equal to that (even) of rest. In fact, if the 
motion is sufficiently rapid, and the fluid is inelastic (a liquid), 
a vacuum may he maintained behind the plate, in which case 
there is evidently no pressure on that side of the plate. 

Whatever pressure exists on the back acts, of course, to 
diminish the resultant resistance. The water on turning the 
sharp corners of the plate is broken up into eddies forming a 



su 



MECHANICS OF ENGINEEKING. 



" wake" behind. From the accompaniment of these eddies, 
the resistance in this case (at least the component JV normal to 
plate) is said to be due to "eddy-making;" though logically 
we should say, rather, that the body does not derive the assist- 
ance (or negative resistance) from behind which it would ob- 
tain if eddies were not formed ; i.e., if the fluid could close in 
behind in smooth curved stream-lines symmetrical with those 
in front. 

The heat corresponding to the change of temperature pro- 
duced in the portion of fluid acted on, by the skin-friction 
and by the mutual friction of the particles in the eddies, is the 
equivalent of the work done (or energy spent) by the motive 
force in maintaining the uniform motion (§ 149). (Joule's 
experiments to determine the Mechanical Equivalent of Heat 
were made with paddles moving in water.) 

If the fluid is sea-water, the results of Col. Beaufoy's ex- 
periments are applicable, viz.: 

The resistance, per square foot of area, sustained by a sub- 
merged plate moving normally to itself [i.e., a = 90°] in sea- 
water with a velocity ofv = 10 ft. per second is 112 lbs. He 
also asserts thdXfor other velocities the resistance varies as the 
square of the velocity. This latter fact we would be led to 
suspect from the results obtained in § 568 for the impulse of 
jets; also in § 565 [see eq. (6)]. Also, that when the plate 
moved obliquely to its normal (as in Fig. 638) the resistance 
was nearly equal to {the resistance, at same velocity, when 
a = 90°) X (the sine of the angle a) / also, that the depth of 
submersion had no influence on the resistance. 

Confining our attention to a plate 'moving nor- 
mally to itself, Fig. 639, let F — area of plate, 
y = heaviness (§ 409) of the fluid, v = the uni- 
form velocity of plate, and g = the acceleration 
of gravity (= 32.2 for the foot and second ; 
== 9.81 for the metre and second). Then from 
the analogy of eq. (6), § 565, where velocity c of 
the jet against a stationary plate corresponds to 
the velocity v of the plate in the present case 
moving through a fluid at rest, we may write 




Fig. 639. 



/ 



PLATES MOVING IN FLUIDS. 815 

Resistance of fluid \ _ p _ r jp v* j v normal ) ^ v 

to moving plate \ ~ ~~ ^ ^ 2g~ ' ' ' \ to plate J ' ' ' ^ ' 

And similarly for the impulse of an indefinite stream of fluid 
against a fixed plate ( ~] to velocity of stream), v being the 
velocity of the current, 

Impulse of current ) __ p _ „,-& tf \v normal ) (c> \ 

upon fixed plate ) ~~ "** ^ 2g ' ' ' \ to plate ("••">■ ' 

The 2g is introduced simply for convenience ; since, having 
v given, we may easily find v* -i- 2g from a table of velocity- 
heads ; and also (a ground of greater importance) since the co- 
efficients C and £' which depend on experiment are evidently 
abstract numbers in the present form of these equations (for 
R and P are forces, and Fyv 1 -i- 2g is the weight (force) of 
an ideal prism of fluid; hence C and C' must be abstract 
numbers.) 

From Col. Beaufoy's experiments (see above), we have for 
^ea-water [ft., lb., sec], putting R = 112 lbs., F— 1 sq. ft., 
y = 64 lbs. per cub. ft., and v = 10 ft. per second, 

2 X 32.2 X 112 

^ 1.0 X 64 X 10 2 

Hence in eq. (1) for sea-water, we may put C = 1.13 (with 
y = 64 lbs. per cub. ft.). 

From the experiments of Dubuat and Thibault, Weisbach 
computes that for the plate of Fig. 639, moving through either 
water or air, C = 1.25 for eq. (1), in which the y for air must 
be computed from § 437 ; while for the impulse of water or 
air on fixed plates he obtains £' = 1.86 for use in eq. (2). It 
is hardly reasonable to suppose that C and £' should not be 
identical in value, and Prof. Unwin thinks that the difference 
in the numbers just given must be due to errors of experi- 
ment. The latter value, £' = 1.86, agrees well with equation 
(6) below. For great velocities C and £' are greater for air 
than for water, since air, being compressible, is of greater 
heaviness in front of the plate than would be computed for 



816 MECHANICS OF ENGINEEKI^G. 

the given temperature and barometric height for use in eqs. 
(1) and (2) 

The experiments of Borda in 1763 led to the formula 

P = [0.0031 + 0.00035c]#y 2 .... (3> 

for the total pressure upon a plate moving through the air 
in a direction ~| to its own surface. P is the pressure in 
pounds, c the length of the contour of the plate in feet, and S 
its surface in square feet, while v is the velocity in miles per 
hour. Adopting the same form of formula, Hagen found, 
from experiments in 1873, the relation 

P= [0.002894 + 0.00014tf]£V ... (4) 

for the same case of fluid resistance. 

Hagen's experiments were conducted with great care, but 
like Borda's were made with a "whirling machine," in which 
the plate was caused to revolve in a horizontal circle of only 
7 or 8 feet radius at the end of a horizontal bar rotating about 
a vertical axis. Hagen's plates ranged from 4 to 40 sq. in. in 
area, and the velocities from 1 to 4 miles per hour. 

The last result was quite closely confirmed by Mr. H. Allen 
Hazen at Washington in November 1886, the experiments 
being made with a whirling machine and plates of from 16 to 
576 sq. in. area. (See the American Journal of Science, Oct. 
1887, p. 245.) 

In Thibault's experiments plates of areas 1.16 and 1.531 sq. 
ft. were exposed to direct wind-pressure, giving the formula 

P = 0.00475/SV. ...... (5) 

Recent experiments in France (see P. P. and Eng. Journal, 
Feb. '87), where flat boards were hung from the side of a rail- 
way train run at different velocities, gave the formula 

P = 0.00535#y 2 (6) 

The highest velocity was 44 miles per hour. The magnitude 
of the area did not seemingly affect the relation given. More 



PLATES IN FLUIDS. 817 

extended and elaborate experiments are needed in this field, 
those involving a motion of translation being considered the 
better, rather than with whirling machines, in which " centrif- 
ugal action" must have a disturbing influence. 

The notation and units for eqs. (4), (5), and (6) are the same 
as those given for (3). 

It may be of interest to note that if equation (3) of § 568 be 
considered applicable to this case of the pressure of an un- 
limited stream of fluid against a plate placed at right-angles to 
the current, with F put equal to the area of the plate, we ob- 
tain, after reduction to the units prescribed above for the pre- 
ceding equations and putting a = 90°, 

P = 0.0053#y 2 (7) 

The value y = 0.0807 lbs. per cub. ft. has been used in the 
substitution, corresponding to a temperature of freezing and 
a barometric height of 30 inches. At higher temperatures, 
of course, y would be less, unless with very high barometer. 

569a. Example. — Supposing each blade of the paddle-wheel 
of a steamer to have an area of 6 sq. ft., and that when in the 
lowest position its velocity [relatively to the water, not to the 
vessel] is 5 ft. per second ; what resistance is it overcoming in 
salt water ? 

From eq. (1) of § 569, with C = 1.13 and y = 64 lbs. per 
cubic foot, we have (ft., lb., sec.) 

jg = 1-13X6 X 64X 26 = m4 

2 X 32.2 

If on the average there may be considered to be three pad- 
dles always overcoming this resistance on each side of the 
boat, then the work lost (work of " slip") in overcoming these 
resistances per second (i.e., power lost) is 

Z, = [6X 169.4] lbs. X 5 ft. per sec. = 5082 ft.-lbs. per sec. 

or 9.24 Horse Power (since 5082 -r- 550 = 9.24). 



818 



MECHANICS OF ENGINEERING. 



If, further, the velocity of the boat is uniform and = 20 ft. 
per sec, the resistance of the water to the progress of the boat 
at this speed being 6 X 169.4, i.e. 1016.4 lbs., the power ex- 
pended in actual propulsion is 

Z 2 = 1016.4 X 20 = 20328 ft.-lbs. per sec. 

Hence the power expended in both ways (usefully in propul- 
sion, uselessly in " slip") is 

L 9 +Z 1 = 25410 ft.-lbs. per sec. = 46.2 H. P. 

Of this, 9.24 H. P., or about 20 per cent, is lost in " slip." 



570. Wind-pressure 
on the surface of a 
roof inclined at an 
angle == a with the 
horizontal, i.e., with 
the direction of the 
wind, is usually esti- 
mated according to 
the empirical formula 




Fig. 640. 



(Hutton's) 



p =p' [sin a] [i-84cosa- 1] ? 



(i) 



in which p' = pressure of wind per unit area against a vertical 
surface (1 to wind), and p = that against the inclined plane 
{and normal to it) at the same velocity. For a value of 
p' = 40 lbs. per square foot (as a maximum), we have the 
following values for p, computed from (1) : 



For a= 5° 


10°' 


15° 


20° 


25° 


30° 


35° 


40° 


45° 


50° 


55° 


60° 


p={\bs. sq. ft.) 5. 2 


9.6 


14 


18.3 


22.5 


26.5 


30.1 


33.4 


36.1 


38.1 


39.6 


40. 



Duchemin's formula for the normal pressure per unit-area is 

2 sin 8 a 



p^p 



1 + sin* a 



{$) 



WIND AND SAIL. 



819 



with the same notation as above. Some experimenters in 
London tested this latter formula by measuring the pressure 
on a metal plate supported in front of the blast-pipe of a blow- 
ing engine ; the results were as follows : 



a = 


15° 


20° 


60° 


90° 


p by experiment = (in lbs. per sq. ft.) 


1.65 


2.05 


3.01 


3.31 


By Ducneinin's formula^? = 


1.60 


2.02 


3.27 


3.31 



The scale of the Smithsonian Institution at Washington for 
the estimation and description of the velocity and pressure of 
the wind is as follows : 



Grade. 


Velocity in 
miles per hour. 


Pressure per 
sq. foot in lbs. 


Name. 








0.00 


Calm. 


1 


2 


0.02 


Very light breeze. 


2 


4 


0.08 


Gentle breeze. 


3 


12 


0.75 


Fresh wind. 


4 


25 


3.00 


Strong wind. 


5 


35 


6 


High wind. 


6 


45 


10 


-Gale. 


7 


60 


18 


Strong gale. 


8 


75 




Violent gale. 


9 


90 




Hurricane. 


10 


100 




Most violent hurricane. 



571. Mechanics of the Sail-boat. — The action of the wind on a 
sail will be understood from the following. Let Fig. 641 
represent the boat in horizontal projection and OS the sail, O 
being the mast. For / ' / , > 

simplicity we consider 
the sail to be a plane 
and to remain vertical. 
At this instant the boat 
is moving in the direc- 
tion M. B of its fore-and- 
aft line with a velocity 
= c, the wind having a velocity of the direction and magni- 
tude represented by w (purposely taken at an angle < 90° with 
the direction of motion of the boat). We are now to inquire 
the nature of the action of the wind on the boat, and whether 




Fig. 641. 



820 MECHANICS OF ENGINEERING. 

in the present position its tendency is to accelerate, or retard, 
the motion of the boat. If we form a parallelogram of which 
w is the diagonal and c one side, then the other side OK, mak- 
ing some angle a with BM, will be the velocity v of the wind 
relatively to the boat (and sail), and upon this (and not upon w) 
depends the action on the sail. The sail, being so placed that 
the angle 6 is smaller than a, will experience pressure from 
the wind ; that is, from the impact of the particles of air which 
strike the surface and glance along it. This pressure, B, is 
normal to the ,sail (considered smooth), and evidently, for the 
position of the parts in the figure, the component of B along 
MB points in the same direction as e, and hence if that com- 
ponent is greater than the water-resistance to the boat at this 
velocity, c will be accelerated; if less, c will be retarded. 
Any change in c, of course, gives a different form to the 
parallelogram of velocities, and thus the relative velocity v 
and the pressure P, for a given position of the sail, will both 
change. [The component of B ~] to JIB tends, of course, to 
cause the boat to move laterally, but the great resistance to 
such movement at even a very slight lateral velocity will make 
the resulting motion insignificant.] 

As c increases, a diminishes, for a given amount and position 
of w ; and the sail must be drawn nearer to the line MB, i.e. 
must be made to decrease, to derive a wind-pressure having 
a forward fore-and-aft component ; and that component be- 
comes smaller and smaller. But if the craft is an ice-boat, this 
small component may still be of sufficient magnitude to exceed 
the resistance and continue the acceleration of c until c is 
larger than w ; i.e., the boat may be caused to go as fast as, or 
faster than, the wind, and still be receiving from the latter a 
forward pressure which exceeds the resistance. And it is 
plain that there is nothing in the geometry of the figure to 
preclude such a relation (i.e., c> w, with 6 < a and > 0). 

572. Resistance of Still Water to Moving Bodies, Completely 
Immersed. — This resistance depends on the shape, position, and 
velocity of the moving body, and also upon the roughness of 
its surface. If it is pointed at both ends (Fig. 642) with its 






MOVING BODIES, IMMERSED. 



821 




axis parallel to the velocity, v, of its uniform motion, the 

stream-lines on closing to- 

gether smoothly at the hinder -— ~:H:irrr~— — -'— 

extremity, or stern, B, exert " 
normal pressures against the 
surface of the portion CD...B 

whose longitudinal compo- - ^_v— ~ zzzzzz-- "_'_'_"_ 

nents approximately balance -— ~~~ 

the corresponding components Fie. 642. 

of the normal pressures on CD . . . A ; so that the resistance 
R, which must be overcome to maintain the uniform velocity 
v, is mainly due to the " skin-friction" alone, distributed along 
the external surface of the body ; the resultant of these resist- 
ances is a force R acting in the line AB of symmetry (sup- 
posing the body symmetrical about the direction of motion). 
If, however, Fig. 643, the stern, E..B ..F is too bluff, 

eddies are formed round the corners 

jB'andi^and the pressure on the 
surface F . . . F is much less than 
in Fig. 642; i.e., the water pres- 
sure from behind is less than the 
backward (longitudinal) pressures 
from in front, and thus the resultant 
resistance R is due partly to skin- 
friction and partly to "eddy-making" (§ 569). 

[Note. — The diminished pressure on FF is analogous to the 
loss of pressure of water (flowing in a pipe) after passing a nar- 
row section the enlargement 
from which to the original 
section is sudden. E.g., Fig. 
644, supposing the velocity v 
and pressure p (per unit-area) 
to be the same respectively 
at A and A\ in the two 
pipes shown, with diameter 
AL = WK=A'L' = WR' 
then the pressure at M i» 
equal to that at A (disregarding skin-friction), whereas that at 




Fig. 643. 




822 



MECHANICS OF ENGINEERING. 



M f is considerably less than that at A' on account of the head 
lost in the sudden enlargement. (See also Fig. 575.)] 

It is therefore evident that bluffness of stem increases the 
resistance much more than bluffness of bow. 

In any case experiment shows that for a given body sym- 
metrical about an axis and moving through a fluid (not only 
water, but any fluid) in the direction of its axis with a uni- 
form velocity = v, we may write approximately the resistance 



R = (resistance at vel. v) = QFy — . 



(i) 



As in preceding paragraphs, F = area of the greatest section, 
~| to axis, of the external surface of body (not of the sub- 
stance) ; i.e., the sectional area of the circumscribing cylinder 
(cylinder in the most general sense) with elements parallel to 
the axis of the body, y — the heaviness (§ 409) of the fluid, 
and v = velocity of motion ; while C is an abstract number 
dependent on experiment. 

According to Weisbach, who cites different experimenters, 
we can put for spheres, moving in water, Q =. about 0.55 ; for 
cannon-balls moving in water, C == .467. 

According to Robins and Hutton, for spheres in air, we 
have 



For v in mets. ) + 
per sec. \ 


5 


25 


100 


200 


300 


400 


KOo j metres 
5UU 1 per sec. 


C=.59 


.63 


.67 


.71 


.77 


.88 


.99 


1.04 



For musket-balls in the air, Piobert found 

C == 0.451 (1 + 0.0023 x veloc. in metres per sec). 
From I>ubuat ? s experiments, for the resistance of water to a 
right prism moving endwise and of length = Z, 



For a: \/F)= 


1 


2 


3 


£ = 1.25 


1.26 


1.31 


1.33 



For a circular cylinder moving perpendicularly to its axis 
Borda claimed that C is one-half as much as for the circum- 



MOVING BODIES, IMMERSED. 823 

scribing right parallelopiped moving with four faces parallel 
to direction of motion. 

Example. — The resistance of the air at a temperature of 
freezing and tension of one atmosphere to a musket-ball \ inch 
in diameter when moving with a velocity of 328 ft. per sec. 
is thus determined by Piobert's formula, above : 

C = 0.451(1 + .0023 X 100) = 0.554 ; 

hence, from eq. (1), 

R = 0.554 X ~ Q|) 2 X -0807 X { ^j = 0.1018 lbs. 

572a. Deviation of a Spinning Ball from a Vertical Plane in 
Still Air. — It is a well-known fact in base-ball playing that if a 
rapid spinning motion is given to the ball about a vertical axis 
as well as a forward motion of translation, its path will not 
remain in its initial vertical plane, but curve out of that plane 
toward the side on which the absolute velocity of an external 
point of the ball's surface is least. Thus, if the ball is thrown 
from North to South, with a spin of such character as to ap- 
pear " clock-wise" seen from above, the ball will curve toward 
the West, out of the vertical plane in which it started. 

This could not occur if the surface of the ball were perfectly 
smooth (there being also no adhesion between that surface and 
the air particles), and is due to the fact that the cushion of com- 
pressed air which the ball piles up in front during its progress, 
and which would occupy a symmetrical position with respect 
to the direction of motion of the centre of the ball if there 
were no motion of rotation of the kind indicated, is now piled 
up somewhat on the East of the centre (in example above), 
creating constantly more obstruction on that side than on the 
right ; the cause of this is that the absolute velocity of the sur- 
face-points, at the same level as the centre of ball, is greatest, 
and the friction greatest, at the instant when they are passing 
through their extreme Easterly positions; since then that 
velocity is the sum of the linear velocity of translation and 
that of rotation ; whereas, in the position diametrically oppo- 



824 MECHANICS OF ENGINEERING. 

site, on the West side, the absolute velocity is the difference / 
hence the greater accumulation of compressed air on the left 
(in the case above imagined, ball thrown from North to South, 
etc.). 

573. Robinson's Cup-anemometer. — This instrument, named 
after Dr. T. R. Eobinson of Armagh, Ireland, consists of four 
hemispherical cups set at equal intervals in a circle, all facing 
in the same direction round the circle, and so mounted on a 
light but rigid framework as to be capable of rotating with 
but little friction about a vertical axis. When in a current of 
air (or other fluid) the apparatus begins to rotate with an ac- 
celerated velocity on account of the pressure against the open 
mouth of a cup on one side being greater than the resistance 
met by the back of the cup diametrically opposite. Yery soon, 
however, the motion becomes practically uniform, the cup- 
centre having a constant linear velocity v" the ratio of which 
to the velocity, v', of the wind at the same instant must be 
found in some way, in order to deduce the value of the latter 
from the observed amount of the former in the practical use 
of the instrument. After sixteen experiments made by Dr. 
Robinson on stationary cups exposed to winds of varying in- 
tensities, from a gentle breeze to a hard gale, the conclusion 
was reached by him that with a given wind- velocity the total 
pressure on a cup with concave surface presented to the wind 
was very nearly four times as great as that exerted when the 
convex side was presented, whatever the velocity (see vol. 
xxn of Transac. Irish Royal Acad., Part 7, p. 163). 

Assuming this ratio to be exactly 4.0 and neglecting axle- 
friction, we have the data for obtaining an approximate value 
of m, the ratio of v' to the observed v", when the instrument is 
in use. The influence of the wind on those cups the planes of 
whose mouths are for the instant || to its direction will also be 
neglected. 

If, then, Fig. 645, we write the impulse on a cup when the 
hollow is presented to the wind [§ 572, eq. (1)] 

*=wH£. cd " 



ROBINSOlSr'S CUP-ANEMOMETER. 825 

and the resistance when the convex side is presented 



v: 




P e = C c Fy^, (2) 

we may also put 

C* = 4£ (S) 

In (1) and (2) v x and v a are relative velocities. 
Kegarding only the two cups A and B, 

whose centres at a definite instant are mov- ►»' 

ing in lines parallel to the direction of the v_ 

wind, it is evident that the motion of the - y h 

cups does not become uniform until the rel- 
ative velocity v' — v" of the wind and cup 
A (retreating before the wind) has become -- — tyk" 
so small, and the relative velocity v' -4- v" - 
with which B advances to meet the air- 
particles has become so great, that the im- 
pulse of the wind on A equals the resist- 
ance encountered by B: i.e., these forces, * v ' B 

J ' ' ' Fig. 645. 

P h and P c , must be equal, having equal 

lever-arms about the axis. Hence, for uniform rotary motion, 

' vyZgX.&Vgl, . . . „ 

i.e. [see eq. (3)], 

4 [^ _1 J = [^ + 1 T ; ° r ' ^-l)' = (^ + !)'•■ (5) 

Solving the quadratic for m, we obtain 

m = 3.00 (6) 

That is, the velocity of the wind is about three times that of 
the cup-centre. 

574. Experiments with Robinson's Cup-anemometer. — The 
ratio 3.00 just obtained is the one in common use in connec- 
tion with this instrument in America. Experiments by Mr. 



826 MECHANICS OE ENGINEERING. 

Hazen at Washington in 1886 (Am. Jour. Science, Oct. '87, 
p. 248) were made on a special type devised by Lieut. Gibbon. 
The anemometer was mounted on a whirling machine at the 
end of a 16-ft. horizontal arm, and values for m obtained, with 
velocities up to 12 miles per hour, from 2.84 to 3.06 ; average 
2.94. The cups were 4 in. in diameter and the distance of their 
centres from the axis 6.72 in., these dimensions being those 
usually adopted in America. This instrument was nearly new 
and was well lubricated. 

Dr. Eobinson himself made an extensive series of experi- 
ments, with instruments of various sizes, of which an account 
may be found in the Philos. Transac. for 1878, p. 797 (see 
also the volume for 1880, p. 1055). Cups of 4 in. and also of 
9 in. were employed, placed first at 24 and then at 12 in. from 
the axis. The cup-centres revolved in a (moving) vertical 
plane perpendicular to the horizontal arm of a whirling- 
machine ; this arm, however, was only 9 ft. long. A friction- 
brake was attached to the axis of the instrument for testing the 
effect of increased friction on the value of in. At high speeds 
of 30 to 40 miles per hour (i.e., the speed of the centre of the 
instrument in its horizontal circle, representing an equal speed 
of wind for an instrument in actual use with axis stationary) 
the effect of friction was relatively less than at low velocities. 
That is, at high speeds with considerable friction the value of 
m was nearly the same as with little friction at low speeds. 
With the large 9 in. cups at a distance of either 24 or 12 in. 
from the axis the value of m at 30 miles per hour ranged 
generally from 2.3 to 2.6, with little or much friction ; while 
with the minimum friction m rose slowly to about 2.9 as the 
velocity diminished to 10 miles per hour. At 5 miles per 
hour with minimum friction m was 3.5 for the 24 in. instru- 
ment and about 5.0 for the 12 in. The effect of considerable 
friction at low speeds was to increase m, making it as high as 
8 or 10 in some cases. With the 4 in.-cups no value was ob- 
tained for m less than 3.3. On the whole, Dr. Robinson con- 
cluded that m is more likely to have a constant value at all 
velocities the larger the cups, the longer the arms, and the less 
the friction, of the anemometer. But few straight-line experi- 



VARIOUS ANEMOMETERS. 827 

ments have been made with the cup-aneinometer, the most 
noteworthy being mentioned on p. 308 of the Engineering 
News for October 1887. The instrument was placed on the 
front of the locomotive of a train running between Baltimore 
and Washington on a calm day. The actual distance is 40 
miles between the two cities, while from the indications of the 
anemometer, assuming m = 3.00, it would have been in one trip 
46 miles and iu another 47. The velocity of the train was 20 
miles per hour in one case and 40 in the other. 

575 Other Anemometers. — Both Biram's and Castello's ane- 
mometers consist of a wheel furnished with radiating vanes 
set obliquely to the axis of the wheel, forming a small " wind- 
mill," somewhat resembling the current-meter for water shown 
in Fig. 604 ; having six or eight blades, however. The wheel 
revolves with but little friction, and is held in the current of 
air with its axis parallel to the direction of the latter, and very 
quickly assumes a steady motion of rotation. The number of 
revolutions in an observed time is read from a dial. The in- 
struments must be rated by experiment, and are used chiefly 
in measuring the velocity of the currents of air in the galleries 
of mines, of draughts of air in flues and ventilating shafts, etc. 

To quote from vol. v of the Report of the Geological Sur- 
vey of Ohio, p. 370 : " Approximate measurements (of the 
velocity of air) are made by miners by flashing gunpowder, 
and noting with a watch the speed with which the smoke 
moves along the air-way of the mine. A lighted lamp is 
sometimes used, the miner moving along the air-gallery, and 
keeping the light in a perfectly perpendicular position, noting 
the time required to pass to a given point." 

Another kind makes use of the principle of Pitot's Tube 
(p. 751), and consists of a U-tube partially filled with water, 
one end of the tube being vertical and open, while the other 
turns horizontally, and is enlarged into a wide funnel, whose 
mouth receives the impulse of the current of air; the differ- 
ence of level of the water in the two parts of the U is a meas- 
ure of the velocity. 



828 MECHANICS OF ENGINEERING. 

576. Resistance of Ships. — We shall first suppose the ship to 
be towed at a uniform speed ; i.e., to be without means of self- 
propulsion (under water). This being the case, it is found that 
at moderate velocities (under six miles per hour), the ship 
"being of "fair" form (i.e., the hull tapering both at bow and 
stern, under water) the resistance in still water is almost wholly 
due to skin-friction, " eddy-making" (see § 569) being done 
away with largely by avoiding a bluff stern. 

"When the velocity is greater than about six miles an hour 
the resistance is much larger than would be accounted for by 
skin-friction alone, and is found to be connected with the sur- 
face-disturbance or waves produced by the motion of the hull 
in (originally) still water. The recent experiments of Mr. 
Froude and his son at Torquay, England, with models, in a tank 
300 feet long, have led to important rules (see Mr. White's 
Naval Architecture and " Hydromechanics" in the Ency. 
Britann.) of so proportioning not only the total length of a 
ship of given displacement, but the length of the entrance (for- 
ward tapering part of hull) and length of run (hinder tapering 
part of hull), as to secure a minimum "wave-making resist- 
ance" as this source of resistance is called. 

To quote from Mr. White (p. 460 of his Naval Architecture, 
London, 1882): "Summing up the foregoing remarks, it 
appears : 

" (1) Th&t frictional resistance, depending upon the area of 
the immersed surface of a ship, its degree of roughness, its 
length, and (about) the square of its speed, is not sensibly 
affected by the forms and proportions of ships; unless there 
be some unwonted singularity of form, or want of fairness. 
For moderate speeds this element of resistance is by far the 
most important ; for high speeds it also occupies an important 
position — from 50 to 60 per cent of the whole resistance, 
probably, in a very large number of classes, when the bottoms 
are clean ; and a larger percentage when the bottoms become 
foul. 

"(2) That eddy-making resistance is usually small, except in 
special cases, and amounts to 8 or 10 per cent of the frictionaJ 



KESISTANCE OF SHIPS. 829 

resistance. A defective form of stern causes largely increased 
eddy-making. 

" (3) That wave-making resistance is the element of the 
total resistance which is most influenced by the forms and pro- 
portions of ships. Its ratio to the frictional resistance, as well 
as its absolute magnitude, depend on many circumstances ; the 
most important being the forms and lengths of the entrance 
and run, in relation to the intended full speed of the ship. 
For every ship there is a limit of speed beyond which each 
small increase in speed is attended by a disproportionate in- 
crease in resistance ; and this limit is fixed by the lengths of 
the entrance and run — the ' wave-making features ' of a ship. 

" The sum of these three elements constitutes the total re- 
sistance offered by the water to the motion of a ship towed 
through it, or propelled by sails ; in a steamship there is an 
6 augment ' of resistance due to the action of the propel- 
lers." 

In the case of a steamship driven by a screw propeller, this 
augment to the resistance varies from 20 to 45 per cent of the 
" tow-rope resistance," on account of the presence and action 
of the propeller itself ; since its action relieves the stern of 
some of the forward hydrostatic pressure of the water closing 
in around it. Still, if the screw is placed far back of the stern, 
the augment is very much diminished ; but such a position in- 
volves risks of various kinds and is rarely adopted. 

We may compute approximately the resistance of the water 
to a ship propelled by steam at a uniform velocity v, in the 
following manner : Let L denote the power developed in the 
engine cylinder ; whence, allowing 10 per cent of L for engine 
friction, and 15 per cent for " work of slip" of the propeller- 
blade, we have remaining 0.75Z, as expended in overcoming 
the resistance R through a distance = v each unit of time ; i.e., 

(approx.) 0.75Z = Rv (1) 

Example.— If 3000 indicated H. P. (§ 132) is exerted by the 
engines of a steamer at a uniform speed of 15 miles per hour 



830 MECHANICS OF E1STGINEEKING. 

(= 22 ft. per sec), we have (with above allowances for slip and 
engine friction) [foot-lb.-sec] 

| X 3000 X 550 = R X 22 ; /.. R = 56250 lbs. 

Further, since R varies (roughly) as the square of the veloc- 
ity, and can therefore be written R = (Const.) X v 2 , we have 
from (1) 

L = a constant X v* ...... (2) 

as a roughly approximate relation between the speed and the 
power necessary to maintain it uniformly. In view of eq. (3) 
involving the cube of the velocity as it does, we can understand 
why a large increase of power is necessary to secure a propor, 
tionally small increase of speed. 

577. " Transporting Power," or Scouring Action, of a Current, 
— The capacity or power of a current of water in an open 
channel to carry along with it loose particles, sand, gravel, 
pebbles, etc., lying upon its bed was investigated experiment 
tally by Dubuat about a century ago, though on a rather small 
scale. His results are as follows : 

The velocity of current must be at least 

0.25 ft. per sec, to transport silt ; 

0.50 " " " loam; 

1.00 " " " sand; 

2.00 " " " gravel; 

3.5 " " " pebbles 1 in. in diam. ; 

4.0 " " " broken stone ; 

5.0 " " " chalk, soft shale. 

However, more modern writers call attention to the fact 
that in some instances beds of sand are left undisturbed by 
currents of greater velocity than that above indicated for sand, 
and explain this fact on the theory that the water-particles 
may not move parallel to the bed, but in cycloids, approxi- 
mately, like the points in the rim of a rolling wheel, so as to 
have little or no scouring action on the bed in those cases. 

In case the particles move in filaments or stream-lines 
parallel to the axis of the stream the statement is sometimes 
made that the " transporting power" varies as the sixth power 



TRANSPORTING POWER" OF CURRENTS. 



831 




Fig. 646. 



of the velocity of the current, by which is meant, more defi- 
nitely, the following : Fig. 64:6. Conceive a row of cubes (or 
other solids geometri- 
cally similar to each 
other) of many sizes, 
all of the same heavi- 
ness (§ 7), and simi- 
larly situated, to be 
placed on the horizon- 
tal bottom of a trough 
and there exposed to 
a current of water, ™' 
being completely im- 
mersed. Suppose the coefficient of friction between the cubes 
and the trough-bottom to be the same for all. Then, as the 
current is given greater and greater velocity v, the impulse 
P m (corresponding to a particular velocity v m ) against some 
one, m, of the cubes, will be just sufficient to move it, and at 
some higher velocity v n the impulse P n against some larger 
cube, n, will be just sufficient to move it, in turn. We are to 
prove that P m : P n : : v m e : v n \ 

Since, when a cube barely begins to move, the impulse is 
equal to the friction on its base> and the frictions under the 
cubes (when motion is impending) are proportional to their 
volumes (see above), we have therefore 



P 



(1) 



Also, the impulses on the cubes, whatever the velocity, are pro- 
portional to the face areas and to the squares of the velocities 
(nearly ; see § 572) ; hence 



v'x r 



(2) 



Prom (1) and (2) we have 



— ^ ) l.C.j — 



(3) 



832 



MECHANICS OF ENGINEERING. 



while from (3) and (2) we have, finally, 



P»::v t 



(4) 



Thus we see in a general way why it is that if the velocity 
of a stream is doubled its transporting power is increased 
about sixty -four-fold ; i.e., it can now impel along the bottom 
pebbles that are sixty-four times as heavy as the heaviest which 
it could move before (of same shape and heaviness). 

Though rocks are generally from two to three times as 
heavy as water, their loss of weight under water causes them to 
encounter less friction on the bottom than if not immersed. 

578. Recent Experiments with Fire-hose, Nozzles, etc. (Ad- 
dendum to § 520.) — The very full and careful investigations of 
Mr. J. R. Freeman, hydraulic engineer of Boston, Mass., in this 
line (see Transac. A. S. 0. E., Nov. 1889) furnish the following 
results: By taking piezometer readings at the ends of a portion. 
of fire-hose conducting a steady flow of water, the values of loss 
of pressure due to fluid friction per 100 feet of length could be 
computed ; a careful measurement being also made of the 
diameter of the hose and of the volume of water transmitted in 
an observed time. The table here given presents results applica- 
ble to hose of exactly 2.5 inches diameter, for a delivery of water 
at the rate of 240 gallons per minute (that is, for a velocity in 
the hose of 15.68 ft. per sec). (The value of / has been com- 
puted by the writer.) 



Sample. 



Description. 



Unlined linen hose 

Woven cotton, rubber-lined, 
"Mill Hose" 

Kl -'•*-, cotton, rubber-lined, hose. 

Ditto, but interstices between 
threads well filled 

Wo /en cotton, rubber-lined, 
hose. So well filled with the 
rubber that the inner surface 
remained smooth under pres- 
sure 



Loss of pressure, 

per 100 ft. of 

length, in lbs. per 

sq. in. 


Coefficient/. 
(See § 520.) 


33.2 




0.01045 


25.5 
19.4 




0.00802 
0.00610 


16.0 




0.00508 


14.1 




0.00443 



Velocity of 

water in 

hose. 



15.68 
15.68 



15.68 



15.68 



It was found that with other rates of flow the friction-head 
varied nearly as the square of the velocity. The great importance 
of a smooth interior of hose is well shown by this table. 

A short section of each kind of hose was filled with liquid 
plaster under pressure. After the setting of the plaster the hose 
was removed and photographs taken of the cast, thus conveying, 
a definite idea of the degree of roughness of interior of hose. 



EXPERIMENTS WITH NOZZLES. 



833 



As to nozzles, it was found that the plain conical nozzle gave 
the best results, jets from the ring-nozzles being slightly inferior 
in range. 

By means of a very delicate form of Pitot's tube measurements 
were made of the velocity in different parts of the section of jets, 
near the nozzle, with the interesting result that in " about two- 
thirds the whole distance from centre to circumference the veloc- 
ity remains the same as at centre," and that at ■§• inch from 
the wall of most of the orifices the velocity was only 5$ less than 
at centre of jet. With a jet from a 5-foot length of brass tubing 
1^ inch in diameter and used as a nozzle the velocity fell off 
rapidly for filaments further from the centre; e.g., at half the 
distance from centre to circumference the velocity was 90$ of 
that at the centre, and at the outside edge 60$. Most of the 
nozzles ranged from 1 in. to 1^ in. in diameter of orifice. 

By using these velocity measurements to " gauge" the flow it 

c a 
was found that the relation h' = — - was quite closely borne out 

(within 1$) (see eq. (7)", p. 804). The point of the Pitot tube 
was conically convergent, its extremity being 0.017 in. in external 
diameter and containing an orifice of 0.006 in. diameter. A 
minute passage-way led from the orifice to a Bourdon gauge. 

Based on his experiments, Mr. Freeman gives tables for the 
maximum vertical height, V, and also the maximum horizontal 
range, H, of "good effective fire-streams" delivered from smooth 
conical nozzles of various sizes and with different piezometer 
pressures p (in lbs. p. sq. in. above atmosphere) at the base of 
play-pipe, the gauge being at same level as nozzle. (The dis- 
tances reached by the extreme drops are very much greater with 
the high pressures. Fand H are in feet.) 

The following is a brief synopsis of this table, d is the internal 
diameter of the extremity of nozzle. The maximum horizontal 
range was obtained at an angle of elevation of about 32°. 



Forp = 


10 


20 


40 


60 


80 


100 




V H 


V 


H 


V 


H 


V H 


V H 


V H 


d = % in. 
d = % " 
d = \ " 
d = \% " 
d = M " 
d = M " 


17 ft. 19 

18 21 
18 21 

18 22 

19 22 

20 23 


33 
34 
35 
36 
37 
38 


29 
33 
37 
38 
40 
42 


60 
62 
64 
65 
67 
69 


44 
49 
55 
59 
63 
66 


72 54 

77 61 
79 67 
83 72 
85 76 
87 79 


79 62 
85 70 
89 76 
92 81 
95 85 
97 88 


83 68 
90 76 
96 83 
99 89 
101 93 
103 96 



834 



MECHANICS OF ENGINEERING. 



579. Addendum on the Pelton Water-wheel. — The annexed cut and 
additional details of the test alluded to on p. 810 are taken from circulars 
of the manufacturers, of San Francisco, Cal. 




The water was measured over an iron weir £" thick and 3.042 feet long 
without end contraction. 

The depth was measured by a Boyden hook gauge reading to .001", and 
was .4146 foot. The quantity of water discharged was found to be 2.819 
cubic feet per second — Fteley's formula. The head lost by friction in pipe 
was 1.8 feet, reducing the effective head to 384.7 feet. 

The work done was measured by a Prony brake bearing vertically down 
upon a platform scale and which showed a weight of 200 pounds upon the 
scale-beam when the brake gear was suspended by a cord from a point im- 
mediately above the wheel-shaft. This made a constant minus correction 
of 200 pounds. The friction pulley had a face of 12", and being kept wet 
by a jet of clear cold water, it developed very little heat and ran without 
much jumping. Thirteen tests were made snowing very uniform results, 
the first four of which were as follows : 



Tests. 


Weight shown by- 
scale. 


Net weight 
(—200 lbs.) 


Rev. wheel-shaft 
per min. 




1 
2 
3 
4 


665 
665 
660 
660 


G = 465 
465 
460 
460 


255 
256 
256£ 
Totals 1,022 
Means 255£ 


Gu = 118342 
118575 
117760 
117990 
472667 
118167 



The arm of the Prony brake was 4.775 feet from centre of wheel-shaft to 
point of contact on scale and hence described a circle with a circumference 
of 30 feet. The work done per minute was therefore Gu(2 itr) = (118167 
x 30) or 3,545,000 foot-pounds, equal to 107.4 horse-power. The theoret- 
ical power of the water was (2.819 x 60 x 384.7 x 62.4) or 4,060,253 foot- 
pounds. The useful effect was therefore 87.3 per cent. 



INDEX TO MECHANICS OF SOLIDS. 

(For index to " Hydraulics" see p. xxii.) 



Page. 

Aberration of Light 90 

Absolute Velocity 89 

Abutment-Line 414 

Abutments of Arches 430, 435 

Acceleration, Angular 107 

Acceleration, Linear 49 

Acceleration, Normal 75, 77 

Acceleration of Gravity 51, 160, 179 

Acceleration, Tangential 74, 80 

Action and Reaction 1, 53 

Angular, or Rotary, Motion. . . 107 

Anomalies in Friction 192 

Anti-Derivative, see Preface 

and p. 253 

Anti-Resultant 402 

Anti-Stress-Resuitant 411 

Apparent Weight 78, 79 

Arches Linear 386, 396 

Arches of Masonry 421, 437 

Arch-Ribs 438, 483 

Arch-Ribs, Classification of 458 

Arch-Rib of Three Hinges . 458,460 
Arch-Rib of Hinged Ends 440, 

458, 461 

Arch-Rib of Fixed Ends 439,459,465 
Arch Trass, or Braced Arch.. . 478 

Atwood's Machine 159 

Autographic Testing Machine. 240 
Beams, Rectangular, Compar- 
ative Strength and Stiffness 

272,273,277 

Belting, Pressure of 181 

Belting, slip of 182 

Bent Lever with Friction 173 

Boat-Rowing 160 

Bow's Notation 7.407 

Box-Girder 275,292 

Braced Arch 438,478 

Brake, Prony 158 

Brakes, Railroad ....... 190 

Bridge, Arch !'.'.. 430 

Bridge-Pier 141 

Bridge, Suspension 46 

Bridge Truss, Warren 35 

Buckling of Web-Plates 383 

Built Beams, designing Sec- 
tions of , 295 

Built Columns 378 

Buit, Prof., Citations from. 224, 229 
Butt-Joint 226 



Page. 

Cantilevers 260, 276, 341 

Cantilever, Oblique 353, 356 

Catenary, Common 46 

Catenary Inverted 387 

Catenary, Transformed 395 

Cast Iron 220, 279 

Cast Iron, Malleable 224 

Centre of Gravity, 18, 19, etc.. . .336 

Centre of Oscillation 119 

Centre of Percussion of Rod. .. Ill 

Centrifugal Action 125 

Centrifugal Force 77, 78 

Centripetal Force 77, 78, 79 

Centrobaric Method 24 

Cheval- Vapeur 136 

Chrome Steel 224 

Circle as Elastic Curve 262, 343 

Circular Arc as Linear Arch. . 391 

Closing Line 414 

Coblenz, Bridge at 459, 478 

Columns, Long 363 

Composition of Forces... 4, 8, 31, 38 
Compression of Short Blocks . . 218 

Concurrent Forces 6, 8, 397 

Cone of Friction 168 

Conical Pendulum 78 

Connecting-Rod 59, 69, 70 

Conservation of Energy 156 

Conservation of Momentum. ." .' . 66 
Continuous Girders, by Analysis 

320—332 

Continuous Girders, by 

Graphics 484—514 

Copying-Press 71 

Cords, Flexible 42 

Cords, Rigidity of 192 

Couples 27, 30 

Cover-Plates 226 

Crane 362 

Crank-Shaft, Strength of 314 

Creeping of Belts 186 

Crushing, Modulus of 219, 424 

Curvilinear Motion ... 72 

Dangerous Section 262, 332 

Dash-Pot 158 

Deck-Beam 275 

Deflections, (Flexure) 252—262, 342 

Derivatives, (Elastic curve) 310 

Derived Quantities 2 

Deviating Force 77 



XV111 INDEX TO MECHANICS OF SOLIDS. 

(For index to " Hydraulics" see p. xxii.) 



Diagrams, Strain 209, 241 

Displacement of Point of Arch 

Rib 447 

Dove-Tail Joint 269 

Duchayla's Proof of the Parallelo- 
gram of Forces 4 

Dynamics, Definition 4 

Dynamics of a Rigid Body 105 

Dynamics, of Material point. ... 49 

Dynamometers 157, 158, 159 

Eddy, Prof. , Graphic Methods, 

(See Preface) 426 

Elastic Curve a Circle . . . 262, 34a 
Elastic Curve an Equilibrium 

Polygon 484 

Elastic Curves.. 245, 252—262, 356 
Elastic Curves, the Four ^-De- 
rivatives of 310 

Elasticity-Line 241 

Elasticity, Modulus of 203, 22? 

Elastic Limit 202 

Elevation of Outer Rail on Rail- 
road Curves 78 

Ellipse of Inertia 94 

Ellipsoid of Inertia 104 

Elliptical Beam 340 

Elongation of Wrought-Iron 

Rod 207 

Energy 137 

Energy, Conservation of 156 

Energy, Kinetic... 137, 144, 147, 150 

Energy, Potential. . . 155 

Equations, Homogeneous 2 

Equator, Apparent Weight at the 78 

Equilibrium 4, 33, 39 

Equilibrium Polygon 401,450 

Equilibrium Polygon Through 

Three Points 418,419 

Equivalent Systems 7, 105, 145 

Exaggeration of Vertical Di- 
mensions in Arch-Ribs. . . . 470 
Examples in Flexure. . . . 280—284 

Examples in Shearing 231, 232 

Examples in Tension and 

Compression 222, 223 

Examples in Torsion 241—243 

. Experiments of an English 

Railroad Commission 314 

Experiments of Hodgkinson 

.. 207, 369 

Experiments of Prof. Lanza... 280 
Experiments on Building Stone 424 

Experiments on Columns 378 

Extrados 421 

Euler's Formula for Columns... 364 

Factor of Safety 223 

Falling Bodies 51 

"False Polygons" 497, 501 

Fatigue of Metals 224 



" Fixed Points " 503 

Flexural Stiffness 250 

Flexure ,244—386 

Flexure and Torsion Combined 314 
Flexure, Beams of Uniform 

Strength 335 

Flexure, Common Theory. . . . 244 
Flexure, Eccentric Load.. .256, 301 
Flexure, Elastic Curves in. . 245, 

251,252—262 

Flexure, Examples in 280 — 284 

Flexure, Hydrostatic Pressure 308 

Flexure, Moving Loads 298 

Flexure, Non-Prismatic 

Beams 332, 335 

Flexure of Long Columns 363 

Flexure of PrismaticBeams 

Under Oblique Forces. . .347—362 

Flexure, Safe Loads in 262—284 

Flexure, Safe Stress in 279 

Flexure, Shearing Stress in 

284—295 

Flexure, Special Problems 

in 295—319 

Flexure, Strength in 249 

Flexure, the Elastic Forces 246 

Flexure, the "Moment" 249 

Flexure, the "Shear" 248 

Flexure, Uniform Load. . . 258, 267 

302, 305, 307, 324, 329. 340 

Flow of Solids 212 

Fly- Wheel.. 121, 151 

Fly -Wheel, Stresses in. . . .126, 127 

Force 1 

Force Diagram 400 

Force Polygons S97 

Forces, Concurrent 6, 8 

Forces, Distributed 197 

Forces, Non-Concurrent 6, 31 

Forces, Parallel 13 

Forces, Parallelogram of 4 

Forces, Varieties of 7 

Free Axes 12£ 

Free-Body Method, the 11, 105 

Friction 164—194, 422, 423 

Frictional Gearing 172 

Friction, Anomalies in 192 

Friction Axle 175 

Friction Brake 158 

Friction, Cone of 168 

Friction in Machinery 191 

Friction, Sliding 164—168 

Friction of Pivots 179 

Friction, Rolling 186 

Friction- Wheels 177 

Friction, Work Spent m 149 

General Properties of Materials 

204 

Governor Ball 78 



INDEX TO MECHANICS OF SOLIDS. 
CFor index to " Hydraulics" see p. xxii.) 



XIX 



Graphic Representations of Uni- 

formally Accelerated Motion. . 57 
Graphic Treatment of Arch .... 431 
Gravity, Acceleration of 51, 79, 160 

Gravity, Centre of 18, 336,453 

Gravity- Vertical 453 

Graphical Statics, Elements 

of 397—420 

Graphical Statics of Vertical 

Forces 412—420 

Guide Curve 83 

Gyroscope 132 

Harmonic Motion 58, 81, 117 

Heat Energy 156 

Heaviness, Table, Etc 3 

Height Due to Velocity 52, 84 

Hodgkinson's Formulae for 

Columns 369 

Hon: ogeneous Equations 2 

Hooke's Law 201, 203, 207 

Hooks, Strength of 362 

Horizontal Straight Girders by 

Graphics 479—483 

Horse-Power 136, 239,242 

.[-Beam 275, 292,295,337 

Ice-Boat, Speed of 90 

Impact » 63 

Impact, Loss of Energy in. . . . 141 

Inclined Beam 359 

Inclined Plane 83, 135, 151, 166, 169 

Indicator 159 

Inertia 53 

Inertia of Piston-Rod 59 

Instantaneous Rotation, Axis 

of 112 

Internal Stress, General Prob- 
lem of 205 

Intrados 421 

Isochronal Axes 120 

Isotropes 204 

Kinetic Energy... 137, 144, 147, 150 

Knot, Fixed 43 

Knot, Slip 43 

Lanza, Experiments of Prof.... 280 

Lateral Contraction 211, 229 

Lateral Security of Girders 280, 298 

Lever 18, 71 

Lever, Bent, "With Friction 173, 174 
Linear Arches... 386—396, 417, 425 

Live Loads 298, 430 

Load-Line 413 

Locomotive on Arch 430 

Locomotive on Girder 298 

Locomotive. Parallel-Rod of... 131 

Malleable Cast Iron 224 

Mass., 2, 53 

Material Point. 3 

Mechanical Equivalent of Heai 156 
Mechanics, Definition of 1 



Mechanics, Divisions of 4 

Middle Third 423 

Modulus of Elasticity 203, 227 

Modulus of Resilience 213 

Modulus of Rupture (Flexure) 278 

Modulus of Tenacity 212 

Moduli of Compression 219 

Mohr's Theorem 486 

Moment- Area 485 

Moment of a Force 14 

Moment-Diagram 263 

Moment of Flexure. . . 248, 348, 351 

Moment of Inertia 91 

Moment of Inertia by Graphics 454 
Moment of Inertia of Box- 
Girder 276 

Moment of Inertia of Built 

Beam 296 

Moment of Inertia of Built 

Column 379 

Moment of Inertia of Plane 

Figures 91—98, 249, 274 

Moment of Inertia of Rigid 

Bodies 98, 103 

Moment of Inertia of Truss 478 

Moment of Torsion 236 

Momentum 66 

Mortar 422 

Motion, Curvilinear 72 

Motion, Rectilinear 50 

Motion, Rotary 107 

Moving Loads (Flexure 298 

Naperian Base 183, 357, 387 

Naperian Logarithms 47 

Navier's Principle 422, 436 

Neutral Axis 245, 247, 347, 355 

Neutral Line (see Elastic Curve.) 

Newton's Laws 1, 53 

Non-Concurrent Forces in a 

Plane. 31, 399 

Non-Concurrent Forces in Space 37 

Normal Acceleration 75, 76, 77 

Normal Stress 200 

Oblique Section of Rod in Ten- 
sion 200 

Parabola as Linear Arch 391 

Parabolic Cord 45 

Parabolic Working-Beam. 336, 344 

Parallel Forces 13 

Parallel- Rod of Locomotive.. 131 

Parallelogram of Forces 4 

Parallelogram of Motions 72 

Parallelogram of Velocities .... 72 

Pendulum, Compound 118 

Pendulum, Conical 78 

Pendulum, Cycloidal 80 

Pendulum, Simple Circular 81 

Phoenix Columns 378 

Pier Reactions 404 



XX 



INDEX TO MECHANICS OF SOLIDS. 
(For index to " Hydraulics" see p. xxii.) 



Piers of Arches 430, 435 

Pile-Driving 140 

Pillars (see columns) 

" Pin-and-Square " Columns... 364 

Planet, Velocity of 82 

Polar Moment of Inertia 97, 238 

Pole (in Graphics) 401 

Pole-Distance 416, 417 

Practical Notes 223 

Potential Energy 155 

Power 134 

Power of Motors 153, 157, 158 

Power, Transmission of, by 

Belting , 184 

Power, Transmission of by 

Shafts 238,318 

Principal Axes 104, 129 

Projectile in Vacuo 83, 84—87 

Prony Friction Brake 158 

Pulley 43, 103 

Punching Rivet Holes 229 

Quantity, Kinds of 1 

Quantities, Derived 2 

Radius of Curvature, 75, 76, 

250,353 

Radius of Gyration 91, 92, 115, 

.....313,376 

Rankine's Formula for Col- 
umns 372, 375 

Rays of Force Diagram 401 

Reaction. . , 1, 18, 36, 53, 404 

Reduced Load-Contour 429 

Regulation of Machines 153 

Relative and Absolute Veloci- 
ties 89 

Resilience. . . .204, 213, 237, 251, 313 
Resultant of Parallel Forces, 13, 15 
Resultant of Two or More 

Forces 4, 6 

Reversal of Stress 514 

Rigid Body 4 

Rigidity of Ropes 192 

Riveting for Built Beams 292 

Rivets and Riveted Plates. 225, 292 
Robinson, Prof. , Integration by 357 

Rod in Tension 198, 200 

Rolling Friction 186 

Rolling Motion 130 

Roof Truss 37, 405 

Rotary Motion 68, 107, 129 

Rotation and Translation Com- 
bined , -130, 150 

Rupture 202 

Safe Limit of Stress 202 

Safe Loads in Flexure . . . 262, 284 

St. Louis Bridge 459, 467, 478 

Schiele's ' 'Anti-Friction" Pivots 181 
Set, Permanent... 202, 209, 208, 241 
Shafts 233—239 



Shafts, Non-Circular 239 

Shear Diagram (Flexure) 265 

Shearing 225—232 

Shearing Distortion 227 

Shear, Distribution of in Flex- 
ure 287 

Shearing Forces 7, 225 

Shearing Stress, 

7,200,201, 225, 234,284 

Shear, the First ^-Derivative 

of Moment (Flexure) 264 

Skidding 190 

Slip (of Oar, etc.) 161 

Slope (in Flexure) 253 

Soffit 421 

Spandrel 421 

Special Equilibrium Polygon 

409,424,440 

Specific Gravity 3 

Statics, Definition of 4 

Statics, Graphical. ....... 397—420 

Statics of a Material Point 8 

Statics of a Rigid Body 27 

Statics of Flexible Cords 42 

Steam Engine Problems... 5$, 

.....61, 69, 70,121,131, 151 

Steam Hammer. 138 

Stiffening of Web-Plates 383 

Stone, Strength of 221, 424 

Stress Diagrams for Arch-Ribs. 471 
Stresses Due to Rib Shorten- 
ing 476 

Strain Diagrams 209, 241 

Strains, two kinds only. 196 

Stress 197, 198 

Stress and Strain, Relation 

Between 201 

Stress-Couple 253, 348 

Stress, Normal and Shearing... 200 

Strength of Materials 195 

Stretching of a Prism Under 

its Own Weight 215 

" Sudden " Application of a 

Load 214,255 

Summation of Products by 

Graphics 451 

Suspension Bridge * . 46 

Table for Flexure 279 

Table for Shearing 228 

Tables for Tension and Com- 
pression 221 

Tackle 43 

Temperature Stresses 206, 217, 

222,473 

Tenacity, Moduli of 212 

Testing Machine, Autographic 240 
Theorem of Three Moments. ... 332 

Thrust (in Flexure) 348, 350 

Torsion 233—243 



INDEX TO MECHANICS OF SOLIDS. 

(For index to " Hydraulics" see p. xxii.) 



XXI 



Torsion, Angle of 233 

Torsion Balance 116 

Torsion, Helix Angle in 233 

Torsion, Moment of 236 

Torsional Resilience 237 

Torsional Stiffness 236 

Torsional Strength 235 

Tractrix, The 181 

Transformed Catenary 395 

Transmission of power by 

Belting 184 

Transmission of Power by Shafting 

238, 318 

Translation, Motion of 68, 

105,106, 133, 137 

Trussed Girders 381 

Uniformly Accelerated Motion 

54, 107 

Uniform Motion 48, 107, 129 

Uniform Strength, Beams of.. 335 
Uniform Strength, Solid of, in 

Tension. 216 



Units, Proper Use see § 6, p. 2 

Velocity, Absolute 89 

Velocity, Angular 107 

Velocity, Linear 49 

Velocity, Relative 89 

Virtual" Moment 67 

Virtual Velocities 67 

Voussoir 386, 421 

Warren Bridge Truss 35 

Water, Jets of 87 

Web of I-Beam 274 

Web of I-Beam, Buckling of.. 383 

Web of I-Beam, Shear in 290 

Wedge, with Friction 171 

Weight.. 2, 3,7, 79 

Weight, Apparent. 78, 79 

Wind and Sail-Boat 89, 90 

Work 133, 134, etc. 

Work and Energy in Machines 

146,147, etc. 

Working-Beam 336, 344 

Working Strength 202' 



INDEX TO HYDRAULICS. 

(For index to " Mechanics of Solids" see p. xvii.) 



PAGE 

Absolute Temperature 606 

Absolute Zero 606 

Accumulator, Hydraulic 700 

Adiabatic Change, 621, 629, 631, 636 
Adiabatic Expansion in Com- 
pressed-air Engine 631 

Adiabatic Flow of Gases from 

an Orifice 778 

Air Collecting in Water-pipes, 

731, 736 
Air, Compressed, Transmission 

of 786-790 

Air-compressor 636 

Air-profile 749 

Air-pump, Sprengel's 656 

Air-thermometer 604 

Amplitude of Backwater 772 

Anemometer, Robinson's. . 824,826 

Anemometer, Biram's 827 

Anemometer, Castello's 827 

Angle of Repose 572 

Angular Stability of Ships 597 

Atkinson Gas-engine 642, 643 

Atmosphere as a Unit Pressure, 519 
Augment of Resistance of Screw 

Propeller 829 

Backwater 771 

Ball, Spinning, Deviation from 

Vertical Plane 823 

Balloon 644 

Barker's Mill 672 

Barometers 530 

Barometric Levelling 619 

Bazin, Experiments!". 688 

Beaufoy's Experiments 814 

Bellinger, Prof., Experiments 

with Elbows. 729 

Bends, Loss of Head due to 728 

Bends in Open Channels 770 

3ent Tube, Liquids in . . . . 529 

Bernoulli's Theorem and the 

Conservation of Energy 717 

Bernoulli's Theorem for Gases, 773 
Bernoulli's Theorem, General 

Form 706 

Bernoulli's Theorem, Steady 

Flow without Friction. . .652, 654 



Bernoulli's Theorem with Fric- 
tion 696 

Bidone, Experiments on Jets. . . 803 

Blowing-engine, Test 776 

Borda's Formula. 722 

Bourdon Steam-gauge 532 

Boyle's Law 615 

Bramah Press 526 

Branching Pipes. 736 

Brayton's Petroleum-engine . . . 641 

Buoyant Effort 586 

Buoyant Effort of the Atmos- 
phere 644 

Canal Lock, Time of Filling, 739, 740 

Centigrade Scale 605 

Centre of Buoyancy 586 

Centre of Pressure 546 

Change of State of Gas 610 

Chezy's Formula 714 

Chezy's Formula for Open Chan- 
nels 758 

Church and Fteley, Report on 
Quaker Bridge Dam, 558, 563, 564 

Clearance 627 

Closed Air-manometer 614 

Coal Consumption 643 

Coal, Heat of Combustion 643 

Coefficient of Contraction 659 

Coefficients of Efflux, 661, 712, 

676, 734, 738, 784 
Coefficient of Fluid Friction, 707, 797 

Coefficient of Resistance 704 

Coefficient of Roughness. . .759, 760 
Coefficient of Velocity, 661, 689, 

704, 712, 723, 734 

Collapse of Tubes 538 

Communicating Prismatic Ves- 
sels 739 

Complete and Perfect Contrac- 
tion 676 

Component of Fluid Pressure., 525 
Compressed-air Engine. ....... 631 

Compressed Air, Transmission 

of 786 

Compressibility of Water 516 

Conical Short Tubes 692 

Conservation of Energy and 
xxii 



INDEX TO HYDRAULICS. 

(For index to "Mechanics of Solids'" see p. xrvii.) 
page J 



XXI 11 



Bernoulli's Theorem 717 

Contraction 659 

Contraction, Perfect 676 

Cooling in Sudden Expansion 

of Gas 622 

Critical Temperature of a Vapor, 608 

Cro ton Aqueduct, Slope 749 

Cunningham, Experiments on 

the River Ganges 759 

Cup-anemometer 824, 826 

Cups, Impulse of Jet on . . . 804, 808 
Current, Transporting Power of, 830 

Current-meters 750 

Curved Dams 562 

Cylinders, Thin Hollow, 

Strength 5?8 

Dams, High Masonry 562 

Darcy, Experiments with Pitot's 

Tube 804 

Darcy and Bazin, Experiments 

with Open Channels 758 

Decrease of Tension of Gas 

along a Pipe 793 

Depth of Flotation 592 

Deviation of Spinning Ball from 

Vertical Plane 823 

Diaphragm in a Pipe, Loss of 

Head 724 

Displacement of a Ship 596 

Divergent Tubes, Flow through, 692 
Dividing Surface of Two Fluids, 528 

Diving-bell 617 

Double Floats 750 

Draught of Ships 596 

Dubuat's Experiments, 707, 815, 

822, 830 
Duchemin's Formula for Wind- 
pressure 818 

Duty of Pumping-enginee 644 

Dynamics of Gaseous Fluids, 773-797 

Earth Pressure 572 

Earthwork Dam 566 

Eddy-making Resistance . . 814, 828 

Efficiency 637, 642 

Efflux of Gases 773-797 

Efflux from Steam-boiler 664 

Efflux from Vessel in Motion.. 670 

Efflux into Condenser 665 

Efflux under Water 669 

Egg-shaped Section for Sewers, 765 

Elastic Fluids 516 

Elbows,Loss of Head due to, 727, 729 
Ellis, Book on Fire-streams . . . 716 
Emptying Vessels, Time of. . 737-746 

Engine, Gas- 641 

Engine, Hot-air 639 

Engine, Petroleum- 641 

Engine, Steam- 624 

Enlargement, Sudden, in Pipe, 721 



Equal Transmission of Fluid 

Pressure 524 

Equation of Continuity, 648, 737, 756 
Equation of Continuity for 

Gases 773 

Equation of Continuity for 

Open Channels 756 

Equilibrium of Flotation 590 

Ericsson's Hot-air Engine 640 

E wart's Experiments on Jets. . . 800 

Expanding Steam 625 

Fahrenheit Scale 605 

Fairbairn's Experiments on Col- 
lapse of Tubes 538 

Fanning, Table of Coefficients 

of Fluid Friction 709 

Feet and Meters, Table 677 

Fire-engine Hose, Friction in. .. 716 

Fire-streams, by Ellis 716 

Floating Staff 750 

Flood-gate 553 

Flotation 590 

Flow in Plane Layers 648, 652 

Flow in Open Channels 749 

Flow of Gas in Pipes. .... .786, 790 

Fluid Friction 695, 797, 828 

Fluid Friction, Coefficient for 

Natural Gas 797 

Fluid Pressure, Equal Trans- 
mission of 524 

Force-pump , 667 

Francis' Formula for Overfalls, 687 
Free Surface of Liquid at Rest, 528 

Free Surface a Paraboloid 544 

FreshWater, Heaviness, Table, 518 
Friction-head in Open Channels, 757 

Friction-head in Pipes 699 

Friction, Fluid 695, 797, 828 

Froude's Experiments on Fluid 

Friction 696 

Froude's Experiments on Grad- 
ual Enlargement in Pipes . . . 725 
Froude's Experiments with 

Piezometers 720 

Fteley and Stearns's Experi- 
ments on Overfalls 687 

Fteley and Stearns's Experi- 
ments with Open Channels. . 758 

Gasand Vapor 607 

Gas-engines 641 

Gaseous Fluids 604-645 

Gas, Flow through Short Pipes, 784 
Gas, Flow through Orifices .. 773-784 

Gas, Illuminating 517, 533 

Gas, Natural, Flow in Pipes, 786, 790 

Gas, Steady Flow 773 

Gas, Velocity of Approach 784 

Gases, Definition , . 515 

Gauging of Streams , ■>.. 755 



XXIV 



INDEX TO HYDRAULICS. 

(For index to " Mechanics of Solids" see p. xvii.) 



PAGE 

Gay-Lussac's Law „ . . 609 

Gradual Enlargement in Pipe.. 725 

Granular Materials 572 

Graphic Representation of 

Change of State of Gas. .... . 628 

Head of Water 530 

Heat-engines, Efficiency 642 

Heaviness of Fluids 517 

Heaviness of Fresh Water at 

Different Temperatures 518 

Height Due to Velocity 649 

Height of the Homogeneous 

Atmosphere 620 

Herschel'sVenturi Water-meter, 726 

High Masonry Dams 562 

Hoop-tension 537 

Hose, Rubber and Leather, Fric- 
tion in 716 

Hot-air Engines 639 

Humphreys and Abbot's Survey 

of the Mississippi River... 754, 759 

Hurdy-gurdy, California 809 

Hutton's Formula for Wind- 
pressure 818 

Hydraulic Grade-line 715 

Hydraulic Mean Depth, 698,757, 764 

Hydraulic Press 526 

Hydraulic Radius 698 

Hydraulic Radius for Minimum 

Friction 764 

Hydraulics, Definition 518 

Hydrodynamics .646-832 

Hydrodynamics or Hydrokinet- 

ics, Definition 518 

Hydromechanics, Definition... 519 

Hydrometers 591 

Hydrometric Pendulum 753 

Hydrostatic Pressure 522 

Hydrostatics, Definition 518 

Ice-making Machine 624 

Illuminating Gas 517, 533 

Immersion of Rigid Bodies 586 

Imperfect Contraction 680, 684 

Impulse and Resistance of 

Fluids 797-832 

Impulse of Jet on Vanes 801, 805 

Inclined Short Tubes, Efiiux 

through 691 

Incomplete Contraction 679, 684 

Inelastic Fluids. 516 

Irreg'ilar Shape, Emptying of 

Vessels of 746 

Isothermal Change. . . .615, 629, 639 

Isothermal Expansion 624, 635 

Isothermal Flow of Gas in 

Pipes 790 

Isothermal Flow of Gases 

through Orifices : . . 777 

Jacket of Hot Water 635 



Jackson's Works on Hydraulics, 761 

Jet from Force-pump 667 

Jets, Impulse of 800, 803, 810 

Jets of Water 660, 662, 833 

Joule, Experiment on Flow of 

Gas 782 

Kansas City Water-works; Si- 
phon 731, 736 

Kinetic Energy 672, 718 

Kinetic Energy of Jet 672, 808 

Kinetic Theory of Gases 516, 

606, 622 

Kutter's Diagram 761 

Kutter's Formula 759 

Kutter's Hydraulic Tables 761 

Laminated Flow. 648, 652 

Land-ties 585 

Law of Charles 609 

Levelling, Barometric 619 

Liquefaction of Oxygen 609 

Liquid, Definition 515 

Long Pipes, Flow of Water 

through 710-716 

Loss of Head 698, 703, 721, etc. 

Loss of Head Due to Bends. . . . 728 
Loss of Head Due to Elbows, 

727, 729 
Loss of Head Due to Throttle- 
valves 730 

Loss of Head Due to Valve- 
gates 730 

Manometers 530 

Mariotte's Law 615, 777 

Mechanics of the Sail-boat 819 

Mendelejeff' s Device for Spec- 
ula. 544 

Metacentre of a Ship 599 

Metres and Feet, Table 677 

Mill, Barker's 672 

Minimum Frictional Resistance 

in Open Channel 764, 766 

Mississippi River, Hydraulic 

Survey 754, 759, 770 

Mixture of Gases 618 

Momentum, Principle of 812 

Moving Pistons 524 

Moving Vane, Impulse of Jet on 805 

Napier on Flow of Steam 781 

Natural Gas, Flow in Pipes, 791, 797 

Natural Slope, of Earth 573 

Non-planar Pistons 526 

Notch, Rectangular, Efflux from, 

683, 741 
Obelisk-shaped Vessel; Time of 

Emptying = 744 

Oblique Impact of Jet on Plate, 810 

Open Channels, Flow in 749-797 

Orifices in Thin Plate 658, 773 

Otto Gas-engine 641, 643 



INDEX TO HYDRAULICS. 

(For index to " Mechanics of Solids" see p. xvii.) 



XXV 



Overfall, Emptying Reservoir 

through 741 

Overfall Weirs. . . .677, 683, 688, 756 
Overfall Weirs, Actual Dis- 
charge 683 

Paddle-wheel of a Steamer 817 

Paraboloid as Free Surface 544 

Paraboloidal Vessel 742 

Parallelopipedical Reservoir 

Walls 555 

Pelton Wheel or Hurdy-gurdy, 809 

Pendulum, Hydrometric 753 

Perfect Fluid, Definition 515 

Permanent Gases 605, 608 

Petroleum-engine 641 

Petroleum Pumping 708 

Piezometer 649, 657, 700 

Pipes, Clean 708 

Pipes, Foul..... 708 

Pipes, Thickness of, for 

Strength 538 

Pipes, Tuberculated 708 

Piston Pressures 523 

Pistons, Moving 524 

Pistons, Non-planar. . , < 526 

Pitot's Tube 751, 804, 827 

Plate between Two Levels of 

Water 568,569 

Plates, Impulse of Jets on.. 801, 

805, 810 

Plates Moving in a Fluid 813 

Plates, Resistance in Sea-water. 814 

Pneumatics, Definition 518 

Poisson's Law 621 

Poncelet's Experiments with 830 

Overfalls 677 

Power Required to Propel 

Ships 830 

Pressure, Centre of 546 

Pressure on Bottom of a Vessel, 545 
Pressure on Curved Surfaces. . . 569 

Pressure on Sluice-gate 551 

Pressure per Unit-area 519 

Pressure-energy 717 

Pressure-head 650 

Principle of Momentum 812 

Pyramidal Vessel, Emptying of, 743 
Quaker Bridge Dam, Proposed, 564 

Radian, Definition 544 

Reaction of a Water- jet 798 

Rectangular Orifices 672, 676 

Refrigerator of Hot-air Engine, 640 
Regenerator of Hot-air Engine, 640 
Relative Equilibrium of a 

Liquid 540 

Reservoir of Irregular Shape, 

Emptying of 746 

Reservoir Walls 554-567 

Resistance, Eddy -making. .814, 828 



Resistance, Wave-making 828 

Resistance of Fluid to Moving 

Bodies.. 820 

Resistance of Fluid to Moving 

Plates 814 

Resistance of Ships 828 

Resistance of Still Water to 

Moving Solids 820 

Retaining Walls 572-580 

Righting Couple, of Floating 

Body 597 

Ritchie-Haskell Direction Cur- 
rent-meter 750 

Rivers, Flow in 749-797 

Robinson, Prof., Experiments 

on Flow of Natural Gas 797 

Robinson's Anemometer 824, 826 

Rounded Orifice 663 

Safety-valves 534 

Sail-boat, Mechanics of the 819 

Sail-boatjfMoving Faster than the 

Wind 820 

Saturated Steam, Heaviness . . . 628 

Saturated Vapor. 607 

Scouring Action of a Current. . 830 
Screw Propeller, Augment of. . 829 

Sewers, Flow in. 761, 765 

Ships, Resistance of 828 

Ships, Stability of 597, 599 

Short Cylindrical Pipes, Efflux 

through 689, 704 

Short Pipes 722, 723 

Short Pipe, Minimum Head for 

Full Discharge 724 

Simpson's Rule 603, 747, 748 

Siphons 735 

Skin-friction 695, 828 

"Slip" 829 

Slope, in Open Channels 749 

Smith, Mr. Hamilton; Hydrau- 
lics 689 

Smithsonian Scale of Wind- 
pressures 819 

Solid of Revolution, Impulse of 

Jet on 803 

Specific Gravity 589 

Sprengel's Air-pump 656 

St. Gothard Tunnel, Experi- 
ments in 787 

Stability of Rectangular Wall. . 554 

Stability of Ships 597 

State of Permanency of Flow. . . 647 

Steady Flow, Definition 647 

Steady Flow, Experimental 

Phenomena 646 

Steady Flow of a Gas 773 

Steam, Expanding 624 

Steam. Flow of 781 

Steam, Saturated, Heaviness of. . 628 



XXVI 



INDEX TO HYDRAULICS. 

(For index to " Mechanics of Solids" see p. xvii.) 



Steam-engine, Examples 627 

Steam-gauge, Bourdon 532 

Stirling's Hot-air Engine 639 

Stream-line 658 

Sudbury Conduit, Experiments, 

758, 762 
Sudden Diminution of Section 

in a Pipe 727 

Sudden Enlargement of Section 

in a Pipe 721 

Surface Floats 750 

Survey, Hydraulic, of Missis- 
sippi River 754, 759 

Tachometer 750 

Temperature, Absolute 606 

Temperature, Influence on 

Flow of Water 703 

Tension of Gas 519 

Tension of Illuminating Gas. .. 533 

Thermodynamics 606 

Thermometers 604 

Thickness of Pipes 538 

Thickness of Pipe for Natural 

Gas 796 

Thin Hollow Cylinders 535 

Thin Plate, Orifices in 658 

Throttle-valves, Loss of Head 

Due to 730 

Time of Emptying Vessels of 

Various Forms 737-746 

Transmission of Compressed 

Air 786, 790 

Transporting Power of a Cur- 
rent 830 

Trapezoidal Section for Open 

Channel 765 

Trapezoidal Wall, Stability of.. 559 

Triangular Orifices 675 

Triangular Wall, Stability 561 

Uniform Motion in Open Chan- 
nel „ 756 

Uniform Rotation of Liquid 542 

Uniform Translation of Liquid.. 540 

Upsetting Couple 599 

Vacuum- chamber, in Siphon. . . 736 
Valve-gates, Loss of Head due 

to 730 

Vanes, Impulse of Jets on. .801, 805 



PAGE. 

Vapors 516, 607 

Variable Diameter, Long Pipe of, 794 
Variable Motion in Open Chan- 
nels 768 

Velocities in Section of River. . 754 
Velocity of Efflux as related to 

Density 668 

Velocity-head 649 

Velocity Measurements in Open 

Channel 750 

Vena Contracta 659 

Venturi's Tube 693 

Venturi Tube, New Forms . 694 

Venturi Water-meter 725 

Volume of Reservoir Found by 
Observing Time of Emptying, 74& 

Water, Compressibility of 516* 

Water-formula for Flow of 

, Gases 774 

Water, Heaviness, at Different 

i Temperatures 518 

Water in Motion 646 

Water-meter, Venturi 725 

Water-ram 530, 538 

Wave-making Resistance 828 

Webb, Prof., Experiments on 

the Reaction of Jets 800 

Wedge of Maximum Thrust. . . 573 
Wedge-shaped Vessel, Time of 

Emptying 742 

Weirs, Overfall. . .677, 683, 688, 

756, 772 

Weisbach's Experiments 682, 

685, 686, 691, 707, 721, 804 
Weisbach's Experiments with 

Elbows and Bends 727, 728 

Weser, River, Backwater in — 772 

Wetted Perimeter 697, 749 

Wex, von, Hydrodynamik 688 

Whirling Machine 816, 826 

Wind-pressure 818 

Wind-pressure, Smithsonian 

Scale of 819 

Woltmann's Mill , 750 

Work of Compressed-air En- 
gine ..•• 631 

Work of Expanding Steam 624 

Work of Jet on a Vane . . . . 807 



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